{"id":22097,"date":"2024-01-10T16:26:17","date_gmt":"2024-01-10T10:56:17","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=22097"},"modified":"2024-01-11T12:40:12","modified_gmt":"2024-01-11T07:10:12","slug":"inter-2nd-year-maths-2a-binomial-theorem-important-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/inter-2nd-year-maths-2a-binomial-theorem-important-questions\/","title":{"rendered":"Inter 2nd Year Maths 2A Binomial Theorem Important Questions"},"content":{"rendered":"

Students get through Maths 2A Important Questions<\/a> Inter 2nd Year Maths 2A Binomial Theorem Important Questions which are most likely to be asked in the exam.<\/p>\n

Intermediate 2nd Year Maths 2A Binomial Theorem Important Questions<\/h2>\n

Question 1.
\nFind the number of terms in the expression of (2x + 3y + z)7<\/sup> [Mar. 14, 13, 07]
\nSolution:
\nNumber of terms in (a + b + c)n<\/sup> are \\(\\frac{(n+1)(n+2)}{2}\\), where n is a positive integer. Hence number of terms in (2x + 3y + z)7<\/sup> are \\(\\frac{(7+1)(7+2)}{2}=\\frac{8 \\times 9}{2}\\) = 36<\/p>\n

Question 2.
\nProve that C0<\/sub> + 2 . C1<\/sub> + 4 . C2<\/sub> + 8 . C3<\/sub> + …………… + 2n<\/sup> . Cn<\/sub> = 3n<\/sup> [A.P. Mar. 15; May 07]
\nSolution:
\nL.H.S. = C0<\/sub> + 2 . C1<\/sub> + 4 . C2<\/sub> + 8 . C3<\/sub> + …………… + 2n<\/sup> . Cn<\/sub>
\n= 0<\/sub> + C1<\/sub>(2) + C2<\/sub> (2)2<\/sup> + C3<\/sub> (23<\/sup>) + …………… + Cn<\/sub> . 2n<\/sup>
\n= (1 + 2)n<\/sup> = 3n<\/sup>
\nNote: (1 + x)n<\/sup> = C0<\/sub> + C1<\/sub> . x + C2<\/sub> . x2<\/sup> + …………… + Cn<\/sub> . xn<\/sup><\/p>\n

Question 3.
\nIf 22<\/sup>Cr<\/sub> is the largest binomial coefficient in the expansion of (1 + x)22<\/sup>, find the value of 13<\/sup>Cr<\/sub>. [A.P. Mar. 15; May 07]
\nSolution:
\nHere n = 22 is an even integer. There is only one largest binomial coefficient and it is
\nn<\/sup>C(n\/2)<\/sub> = 22<\/sup>C11<\/sub> = 22<\/sup>Cr<\/sub> \u21d2 r = 11
\n\u2234 13<\/sup>Cr<\/sub> = 13<\/sup>C11<\/sub> = 13<\/sup>C2<\/sub> = \\(\\frac{13 \\times 12}{1 \\times 2}\\) = 78<\/p>\n

Question 4.
\nWrite down and simplify 6th<\/sup> term in (\\(\\frac{2x}{3}\\) + \\(\\frac{3y}{2}\\))9<\/sup> [May 13]
\nSolution:
\n6th<\/sup> term in (\\(\\frac{2x}{3}\\) + \\(\\frac{3y}{2}\\))9<\/sup>
\nThe general term in (\\(\\frac{2x}{3}\\) + \\(\\frac{3y}{2}\\))9<\/sup> is
\nTr+1<\/sub> = 9<\/sup>Cr<\/sub> (\\(\\frac{2x}{3}\\))9-r<\/sup> (\\(\\frac{3y}{2}\\))r<\/sup>
\nPut r = 5
\nT6<\/sub> = 9<\/sup>C5<\/sub> (\\(\\frac{2x}{3}\\))4<\/sup> (\\(\\frac{3y}{2}\\))5<\/sup>
\n= 9<\/sup>C5<\/sub> (\\(\\frac{2}{3}\\))4<\/sup> (\\(\\frac{3}{2}\\))5<\/sup> x4<\/sup> y5<\/sup>
\n= \\(\\frac{9 \\times 8 \\times 7 \\times 6}{1 \\times 2 \\times 3 \\times 4} \\frac{\\left(2^{4}\\right)}{3^{4}} \\cdot \\frac{3^{5}}{2^{5}} \\cdot x^{4} y^{5}\\)
\n= 189 x4<\/sup>y5<\/sup><\/p>\n

Question 5.
\nIf the coefficients of (2r + 4)th<\/sup> term and (3r + 4)th<\/sup> term in the expansion of (1 + x)21<\/sup> are equal, find r. [T.S. Mar.15]
\nSolution:
\nT2r+4<\/sub> in (1 + x)21<\/sup> is
\n= 21<\/sup>C2r+3<\/sub> (x)2r+3<\/sup> ………………… (1)
\nT3r+4<\/sub> in (1 + x)21<\/sup> is
\n= 21<\/sup>C3r+3<\/sub> . (x)3r+3<\/sup> ……………….. (2)
\n\u21d2 Coefficients are equal
\n\u21d2 21<\/sup>C2r+3<\/sub> = 21<\/sup>C3r+3<\/sub>
\n\u21d2 21 = (2r + 3) + (3r + 3)
\n(or)
\n2r + 3 = 3r + 3
\n\u21d2 5r = 15
\n\u21d2 r = 3 (or) r = 0 .
\nHence r = 0, 3.<\/p>\n

\"Inter<\/p>\n

Question 6.
\nFin the sum of the infinite series 1 + \\(\\frac{1}{3}\\) + \\(\\frac{1.3}{3.6}\\) + \\(\\frac{1.3.5}{3.6.9}\\) + …………… [T.S. Mar. 15]
\nSolution:
\nThe series can be written as
\nS = 1 + \\(\\frac{1}{1}\\). \\(\\frac{1}{3}\\) + \\(\\frac{1.3}{3.6}\\) (\\(\\frac{1}{3}\\))2<\/sup> + \\(\\frac{1.3.5}{1.2.3}\\) (\\(\\frac{1}{3}\\))3<\/sup> + ……………..
\nThe series of the right is of the form
\n1 + \\(\\frac{p}{1}\\) (\\(\\frac{x}{q}\\)) + \\(\\frac{p(p+q)}{1.2}\\) (\\(\\frac{x}{q}\\))2<\/sup> + \\(\\frac{p(p+q)(p+2 q)}{1.2 .3}\\) (\\(\\frac{x}{q}\\))3<\/sup> + ……………
\nHere p = 1, q = 2, \\(\\frac{x}{q}\\) = \\(\\frac{1}{3}\\) \u21d2 x = \\(\\frac{2}{3}\\)
\nThe sum of the given series
\nS = (1 – x)-p\/q<\/sup>
\n= (1 – \\(\\frac{2}{3}\\))-1\/2<\/sup> = (\\(\\frac{1}{3}\\))-1\/2<\/sup> = \\(\\sqrt{3}\\)<\/p>\n

Question 7.
\nFind the set E of the value of x for which the binomial expansions for the (a + bx)r<\/sup> are valid. [Mar. 08]
\nSolution:
\n(4 + 9x)-2\/3<\/sup> = 4-2\/3<\/sup> [1 + \\(\\frac{9x}{4}\\)]-2\/3<\/sup>
\nThe binomial expansion of (4 + 9x)-2\/3<\/sup> is valid
\nWhen |\\(\\frac{9x}{4}\\)| < 1
\n\u21d2 |x| < \\(\\frac{4}{9}\\)
\n\u21d2 x \u2208 (\\(\\frac{-4}{9}\\), \\(\\frac{4}{9}\\))
\ni.e., E = (\\(\\frac{-4}{9}\\), \\(\\frac{4}{9}\\))<\/p>\n

Question 8.
\nIf the 2nd<\/sup>, 3rd<\/sup> and 4th<\/sup> terms in the expansion of (a + x)n<\/sup> are respectively 240, 720,
\n1080, find a, x, n. [T.S. Mar. 16]
\nSolution:
\nT2<\/sub> = 240 \u21d2 n<\/sup>C1<\/sub> an-1<\/sup> x = 240 …………….. (1)
\nT3<\/sub> = 720 \u21d2 n<\/sup>C2<\/sub> an-2<\/sup> x2<\/sup> = 720 …………… (2)
\nT4<\/sub> = 1080 \u21d2 n<\/sup>C3<\/sub> an-3<\/sup> x3<\/sup> = 1080 …………… (3)
\n\\(\\frac{(2)}{(1)} \\Rightarrow \\frac{{ }^{n} C_{2} a^{n-2} x^{2}}{{ }^{n} C_{1} a^{n-1} x}=\\begin{aligned}
\n&720 \\\\
\n&240
\n\\end{aligned}\\)
\n\u21d2 \\(\\frac{n-1}{2} \\frac{x}{a}\\) = 3 \u21d2 (n – 1)x = 6a …………………. (4)
\n\\(\\frac{(3)}{(2)} \\Rightarrow \\frac{{ }^{n} C_{3} a^{n-3} x^{3}}{{ }^{n} C_{2} a^{n-2} x^{2}}=\\frac{1080}{720}\\)
\n\u21d2 \\(\\frac{n-2}{3} \\frac{x}{a}=\\frac{3}{2}\\)
\n\u21d2 2(n – 2)x = 9a …………………… (5)
\n\\(\\frac{(4)}{(5)} \\Rightarrow \\frac{(n-1) x}{2(n-2) x}=\\frac{6 a}{9 a} \\Rightarrow \\frac{n-1}{2 n-4}=\\frac{2}{3}\\)
\n\u21d2 3n – 3 = 4n – 8
\n\u21d2 n = 5
\nFrom (4), (5 – 1) x = 6a \u21d2 4x = 6a
\n\u21d2 x = \\(\\frac{3}{2}\\) a
\nSubstitute x = \\(\\frac{3}{2}\\) a, n = 5 in (1)
\n5<\/sup>C1<\/sub> . a4<\/sup> . \\(\\frac{3}{2}\\) a = 240
\n5 \u00d7 \\(\\frac{3}{2}\\) a5<\/sup> = 240
\na5<\/sup> = \\(\\frac{480}{15}\\) = 32 = 25<\/sup>
\n\u2234 a = 2, x = \\(\\frac{3}{2}\\) a = \\(\\frac{3}{2}\\) (2) = 3
\n\u2234 a = 2, x = 3, n = 5.<\/p>\n

Question 9.
\nIf the coefficients of rth<\/sup>, (r + 1)th<\/sup>, and (r + 2)nd<\/sup>, terms in the expansion of (1 + x)n<\/sup>, are in A.P. then show that n2<\/sup> – (4r + 1)n + 4r2<\/sup> – 2 = 0. [T.S. Mar. 15, 08]
\nSolution:
\nCoefficient of Tr<\/sub> = n<\/sup>Cr-1<\/sub>
\nCoefficient of Tr+1<\/sub> = n<\/sup>Cr<\/sub>
\nCoefficient of Tr+2<\/sub> = n<\/sup>Cr+1<\/sub>
\nGiven n<\/sup>Cr-1<\/sub>, n<\/sup>Cr<\/sub>, n<\/sup>Cr+1<\/sub> are in A.P.
\n\u21d2 2 . n<\/sup>Cr<\/sub> = n<\/sup>Cr-1<\/sub> + n<\/sup>Cr+1<\/sub>
\n\"Inter
\n\u21d2 (2n – 3r + 2) (r + 1) = (n – r) (n – r + 1)
\n\u21d2 2nr + 2n – 3r2<\/sup> – 3r + 2r + 2 = n2<\/sup> – 2nr + r2<\/sup> + n – r
\n\u21d2 n2<\/sup> – 4nr + 4r2<\/sup> – n – 2 = 0
\n\u2234 n2<\/sup> – (4r + 1)n + 4r2<\/sup> – 2 = 0<\/p>\n

Question 10.
\nIf n is a postive integer, prove that \\(\\sum_{r=1}^{n} r^{3}\\left(\\frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}\\right)^{2}=\\frac{(n)(n+1)^{2}(n+2)}{12}\\) [Mar. 13]
\nSolution:
\n\"Inter
\n= (n + 1)2<\/sup> \u03a3 r – 2(n + 1) \u03a3 r2<\/sup> + \u03a3 r3<\/sup>
\n= (n + 1)2<\/sup> \\(\\frac{(n)(n+1)}{2}\\)
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 11.
\nFind the set of values of x for which the binomial expansions of the following are valid.
\n(i) (2 + 3x)-2\/3<\/sup>
\n(ii) (5 + x)3\/2<\/sup>
\n(iii) (7 + 3x)-5<\/sup>
\n(iv) (4 – \\(\\frac{x}{3}\\))-1\/2<\/sup> [A.P. Mar. 17; Mar. 16; Mar. 11]
\nSolution:
\n(i) (2 + 3x)-2\/3<\/sup> = [2(1 + \\(\\frac{3}{2}\\)x)]-2\/3<\/sup>
\n= 2-2\/3<\/sup> (1 + \\(\\frac{3}{2}\\)x)-2\/3<\/sup>
\n\u2234 The binomial expansion of (2 + 3x)-2\/3<\/sup> is valid when |\\(\\frac{3}{2}\\)x| < 1
\n(i.e.,) |x| < \\(\\frac{2}{3}\\)
\n(i.e.,) x \u2208 (-\\(\\frac{2}{3}\\), \\(\\frac{2}{3}\\))<\/p>\n

ii) (5 + x)3\/2<\/sup> = [5 (1 + \\(\\frac{x}{5}\\))]3\/2<\/sup> [T.S. Mar. 17]
\n= 53\/2<\/sup> (1 + \\(\\frac{x}{5}\\))]3\/2<\/sup>
\n\u2234 The binomial expansion of (5 + x)3\/2<\/sup> is valid when \\(\\frac{x}{5}\\) < 1
\n(i.e.,) |x| < 5
\n(i.e.,) x \u2208 (-5, 5)<\/p>\n

iii) (7 + 3x)-5<\/sup> = [7 (1 + \\(\\frac{3}{7}\\) x)]-5<\/sup>
\n= 7-5<\/sup> (1 + \\(\\frac{3}{7}\\) x)]-5<\/sup>
\n(7 + 3x)-5<\/sup> is valid when \\(\\frac{3x}{7}\\) < 1
\n\u21d2 |x| < \\(\\frac{7}{3}\\) \u21d2 x \u2208 (\\(\\frac{-7}{3}\\), \\(\\frac{7}{3}\\))<\/p>\n

iv) (4 – \\(\\frac{x}{3}\\))-1\/2<\/sup> = [4(1 – \\(\\frac{x}{3}\\))]-1\/2<\/sup>
\n(4 – \\(\\frac{x}{3}\\))-1\/2<\/sup> is valid when \\(\\frac{-x}{12}\\) < 1
\n\u21d2 |x| < 12
\n\u21d2 x \u2208 (-12, 12)<\/p>\n

Question 12.
\nFind the sum of the infinite series
\n\\(\\frac{3}{4}\\) + \\(\\frac{3.5}{4.8}\\) + \\(\\frac{3.5 .7}{4.8 .12}\\) + …… (Mar. 11)
\nSolution:
\n\"Inter<\/p>\n

Question 13.
\nIf x = \\(\\frac{1.3}{3.6}\\) + \\(\\frac{1.3 .5}{3.6 .9}\\) + \\(\\frac{1.3 .5 .7}{3.6 .9 .12}\\) + …… then prove that 9x2<\/sup> + 24x = 11 (TS Mar. ’16, AP Mar. ’17, ’15)
\nSolution:
\n\"Inter
\n\u21d2 3x + 4 = 3\\(\\sqrt{3}\\)
\nSquaring on both sides
\n(3x + 4)2<\/sup> = (3\\(\\sqrt{3}\\))2<\/sup>
\n\u21d2 9x2<\/sup> + 24x + 16 = 27
\n\u21d2 9x2<\/sup> + 24x = 11<\/p>\n

Question 14.
\nIf x = \\(\\frac{5}{(2 !) \\cdot 3}\\) + \\(\\frac{5.7}{(3 !) \\cdot 3^{2}}\\) + \\(\\frac{5.7 .9}{(4 !) \\cdot 3^{3}}\\) + …… then find the value of x2<\/sup> + 4x. (mar. 13)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 15.
\nFind the sum of the infinite series
\n\\(\\frac{7}{5}\\) (1 + \\(\\frac{1}{10^{2}}\\) + \\(\\frac{1.3}{1.2}\\).\\(\\frac{1}{10^{4}}\\) + \\(\\frac{1.3 .5}{1.2 .3}\\).\\(\\frac{1}{10^{6}}\\) + …….) (AP Mar. \u201816, May 13; Mar. ’05)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 16.
\nFor n = 0, 1, 2, 3, ….n, prove that
\n\"Inter
\n(TS Mar. ’15)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 17.
\nIf x = \\(\\frac{1}{5}\\) + \\(\\frac{1.3}{5.10}\\) + \\(\\frac{1.3 .5}{5.10 .15}\\) + …… \u221e find 3x2<\/sup> + 6x. (May. ’14, ’07, ’06; May. ’11)
\nSolution:
\nGiven that
\n\"Inter
\n\u21d2 3(1 + x)2<\/sup> = 5
\n\u21d2 3x2<\/sup> + 6x + 3= 5
\n\u21d2 3x2<\/sup> + 6x = 2<\/p>\n

Question 18.
\nWrite the expansion or (2a + 3b)6<\/sup>.
\nSolution:
\n\"Inter<\/p>\n

Question 19.
\nFind the 5th term in the expansion of (3x – 4y)7<\/sup>.
\nSolution:
\nT5<\/sub> = T4 + 1<\/sub>
\n= 7<\/sup>C4<\/sub> (3x)7 – 4<\/sup> (-4y)4<\/sup>
\n= 35.27x3<\/sup>. 256y4<\/sup>
\n= 241920 x3<\/sup> y4<\/sup><\/p>\n

\"Inter<\/p>\n

Question 20.
\nFind the 4th<\/sup> term from the end in the expansion (2a + 5b)8<\/sup>.
\nSolution:
\n(2a + 5b)8<\/sup> expansion contain 9 terms. The fourth term from the end is 6th<\/sup> term from the beginning.
\n\u2234 8<\/sup>C5<\/sub> (2a)8 – 5<\/sup> (5b)5<\/sup>
\n= 8<\/sup>C5<\/sub> (2a)8 – 5<\/sup> (5b)5<\/sup>
\n= 8<\/sup>C5<\/sub> . 23<\/sup>. 55<\/sup> a3<\/sup> b5<\/sup><\/p>\n

Question 21.
\nFind the middle term of the following expansions
\n(i) (3a – 5b)6<\/sup>
\n(ii) (2x + 3y)7<\/sup>
\nSolution:
\ni) Here n = 6 (even)
\n\u2234 \\(\\frac{n}{2}\\) + 1 = \\(\\frac{6}{2}\\) + 1 = 4th<\/sup> term is the middle term
\n\u2234 T4<\/sub> = T3 + 1<\/sub>
\n= 6<\/sup>C3<\/sub> (3a)6 – 3<\/sup> (-5b)3<\/sup>,
\n= –6<\/sup>C3<\/sub>. 33<\/sup>. 53<\/sup>. a3<\/sup> b3<\/sup><\/p>\n

ii) Here n = 7 (odd)
\n\\(\\frac{\\mathrm{n}+1}{2}\\) = \\(\\frac{7+1}{2}\\) = 4, \\(\\frac{\\mathrm{n}+3}{2}\\) = \\(\\frac{7+3}{2}\\) = 5
\n\u2234 4th<\/sup>, 5th<\/sup> terms are middle terms.
\n\u2234 T4<\/sub> = T3 + 1<\/sub> = 7<\/sup>C3<\/sub> (2x)7 – 3<\/sup> (3y)3<\/sup> = 7<\/sup>C3<\/sub> 24<\/sup> 33<\/sup>.x4<\/sup>.y3<\/sup>
\nT5<\/sub> = T4 + 1<\/sub> = 7<\/sup>C4<\/sub>(2x)7-4<\/sup>(3y)4<\/sup> = 7<\/sup>C4<\/sub>. 23<\/sup>.34<\/sup>. x3<\/sup>. y4<\/sup><\/p>\n

Question 22.
\nn is a positive integer then prove that
\nSolution:
\ni) Co<\/sub> + C1<\/sub> + C2<\/sub> + …….. + Cn<\/sub> = 2n<\/sup>
\nii) a) Co<\/sub> + C2<\/sub> + C4<\/sub> + … + Cn<\/sub> = 2n – 1<\/sup> if n is even
\n(b) Co<\/sub> + C2<\/sub> + C4<\/sub> + …. + Cn – 1<\/sub> = 2n – 1<\/sup> if n is odd.
\niii) (a) C1<\/sub> + C3<\/sub> + C5<\/sub> + …. + Cn – 1<\/sub> = 2n – 1<\/sup> if n is even.
\n(b) C1<\/sub> + C3<\/sub> + C5<\/sub> + …. + Cn – 1<\/sub> = 2n – 1<\/sup> if n is odd.
\nSolution:
\nWe know (1 + x)n<\/sup> = n<\/sup>C0<\/sub> + n<\/sup>C1<\/sub> x + n<\/sup>C2<\/sub> x2<\/sup>+ …… + n<\/sup>Cn<\/sub> xn<\/sup>
\n= C0<\/sub> + C1<\/sub>x + C2<\/sub>x2<\/sup> + …… + Cn<\/sub>xn<\/sup><\/p>\n

\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 23.
\nProve that C0<\/sub> + 3.C1<\/sub> + 5.C2<\/sub> + ……… +(2n + 1). Cn<\/sub> = (2n + 2). 2n – 1<\/sup>.
\nSolution:
\nLet S = C0<\/sub> + 3.C1<\/sub> + 5.C2<\/sub> + …… + (2n + 1). Cn<\/sub> —— (1)
\nBy writing the terms in (1) in the reverse older, we get
\n\"Inter<\/p>\n

Question 24.
\nFind the numerically greatest term in the binomial expansion of (1 – 5x)12<\/sup> when x = \\(\\frac{2}{3}\\).
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 25.
\nCompute numerically ireatist term (s) in the expansionly of (3x – 5y)n<\/sup> when x = \\(\\frac{3}{4}\\), y = \\(\\frac{2}{7}\\) and n = 17
\nSolution:
\nGiven x = \\(\\frac{3}{4}\\), y = \\(\\frac{2}{7}\\) and n = 17
\n\"Inter
\n\"Inter<\/p>\n

Question 26.
\nFind the largest binomial coefficients (s) in the expansion of
\n(i) (1 + x)19<\/sup>
\n(ii) (1 + x)24<\/sup>
\nSolution:
\n(i) Here n = 19 is an odd integer. Hence the largest binomial coefficients are
\n\"Inter
\n(ii) Here n = 24 is an even integer. Hence the largest binomial coefficient is
\n\"Inter<\/p>\n

Question 27.
\nIf 22<\/sup>Cr<\/sub> is the largest binomial coefficient in the expansion of (1 + x)22<\/sup>, find the value of 13<\/sup>Cr<\/sub>. (A.P. Mar\u201916, May \u201811)
\nSolution:
\nHere n = 22 is an even integer. There is only one largest binomial coefficient and it is
\n\"Inter<\/p>\n

Question 28.
\nFind the 7th<\/sup> term in the expansion of \\(\\left(\\frac{4}{x^{3}}+\\frac{x^{2}}{2}\\right)^{14}\\)
\nSolution:
\nThe general term in the expansion of
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 29.
\nFind the 3rd<\/sup> term from the end in the expansion of \\(\\left(x^{-2 \/ 3}-\\frac{3}{x^{2}}\\right)^{8}\\)
\nSolution:
\nComparing with (X + a)n<\/sup>, we get
\nX = x-2\/3<\/sup>, a = \\(\\frac{-3}{x^{2}}\\), n = 8
\nIn the given expansion \\(\\left(x^{-2 \/ 3}-\\frac{3}{x^{2}}\\right)^{8}\\), we have n + 1 = 8 + 1 = 9 terms
\nHence the 3rd<\/sup> term from the end is 7th<\/sup> term from the beginning.
\n\"Inter<\/p>\n

Question 30.
\nFind the coefficient of x9<\/sup> and x10<\/sup> in the expansion of \\(\\left(2 x^{2}-\\frac{1}{x}\\right)^{2 c}\\)
\nSolution:
\nIf we write X = 2x2<\/sup> and a = –\\(\\frac{1}{x}\\), then the general term in the expansion of
\n\"Inter
\nSince r = \\(\\frac{31}{3}\\) which is impossible since r must be a positive integer. Thus \u2018there is no term containing x9<\/sup> in the expansion of the given expression. In otherwords the coefficient of x9<\/sup> is \u20180\u2019.
\nNow, to find the coefficient of x10<\/sup>.
\nput 40 – 3r = 10
\n\u21d2 r = 10
\n\"Inter<\/p>\n

Question 31.
\nFind the term independent of x (that is the constant term) in the expansion of
\n\"Inter
\nSolution:
\n\"Inter<\/p>\n

Question 32.
\nIf the coefficients of x10<\/sup> in the expansion of \\(\\left(a x^{2}+\\frac{1}{b x}\\right)^{11}\\) is equal to the coefficient of x-10<\/sup> in the expansion of \\(\\left(a x-\\frac{1}{b x^{2}}\\right)^{11}\\) ; find the relation between a and b where a and b are real numbers.
\nSolution:
\nThe general term in the expansion of \\(\\left(a x^{2}+\\frac{1}{b x}\\right)^{11}\\) is
\n\"Inter
\nTo find the coefficient of x10<\/sup>, put
\n22 – 3r = 10 \u21d2 3r = 12 \u21d2 r = 4
\nHence the coefficient of x10<\/sup> in
\n\"Inter
\nGiven that the coefficients are equal.
\nHence from (1) and (2), we get
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 33.
\nIf the kth<\/sup> term is the middle term in the expansion of \\(\\left(x^{2}-\\frac{1}{2 x}\\right)^{20}\\), find Tk<\/sub> and Tk + 3<\/sub>.
\nSolution:
\nThe general term in the expansion of
\n\"Inter
\n\"Inter<\/p>\n

Question 34.
\nIf the coefficients of (2r + 4)th<\/sup> and (r – 2)nd<\/sup> terms in the expansion of (1 + x)18<\/sup> are equal, find r.
\nSolution:
\n\"Inter<\/p>\n

Question 35.
\nProve that 2.C0<\/sub> + 7.C1<\/sub> + 12.C2<\/sub> + …… + (5n + 2)Cn<\/sub> = (5n + 4)2n – 1<\/sub>.
\nSolution:
\nFirst method:
\nThe coefficients of C0<\/sub>, C1<\/sub>, C2<\/sub>, …., Cn<\/sub> are in A.P. with first term a = 2, C.d (d) = 5
\n\"Inter
\nSecond method:
\nGeneral term in LH.S.
\n\"Inter
\n\"Inter<\/p>\n

Question 36.
\nProve that
\n\"Inter
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 37.
\nFor n = 0, 1, 2, 3 , n, prove that C0<\/sub>. Cr<\/sub> + C1<\/sub>. Cr + 1<\/sub> + C2<\/sub>. Cr + 2<\/sub> + ……. + Cn – r<\/sub>. Cn<\/sub> = 2n<\/sup>Cn + 1<\/sub>. (T.S. Mar. ’15)
\n\"Inter
\nSolution:
\nWe know that
\n\"Inter
\n\"Inter
\n\"Inter<\/p>\n

Question 38.
\nProve that
\n\"Inter
\nSolution:
\n\"Inter<\/p>\n

Question 39.
\nFind the numerically greatest term (s) in the expansion of<\/p>\n

i) (2 + 3x)10<\/sup> when x = \\(\\frac{11}{8}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

ii) (3x – 4y)14<\/sup> when x = 8, y = 3.
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 40.
\nProve that 62n<\/sup> – 35n – 1 is divisible by 1225 for all natural numbers of n.
\nSolution:
\n62n<\/sup> – 35n – 1 = (36)n<\/sup> – 35n – 1
\n= (35 + 1)n<\/sup> – 35n – 1
\n\"Inter
\nHence 62n<\/sup> – 35n – 1 is divisible by 1225 for all integral values of n.<\/p>\n

Question 41.
\nSuppose that n is a natural number and I, F are respectively the integral part and fractional part of (7 + \\(\\sqrt{3}\\))n<\/sup>. Then show that
\n(i) I is an odd integer
\n(ii) (I + F) (I – F) = 1
\nSolution:
\nGiven that (7 + 4\\(\\sqrt{3}\\))n<\/sup> = I + F where I is an integer and 0 < F < 1
\n\"Inter
\n= 2k, where k is a positive integer —— (1)
\nThus I + F + f is n even integer.
\nSince I is an integer, we get that F + f is an integer. Also since 0 < F < 1 and 0 < f < 1
\n\u21d2 0 < F + f < 2
\n\u2235 F + 1 is an integer
\nWe get F + f = 1
\n(i.e.,) I – F = f ——— (2)<\/p>\n

(i) From (1) I + F + f = 2k
\n\u21d2 f = 2k – 1, an odd integer.
\n(ii) (I + F) (I – F) = (I + F) f
\n= (7 + 4\\(\\sqrt{3}\\))n<\/sup> (7 – 4\\(\\sqrt{3}\\))n<\/sup>
\n= (49 – 48)n<\/sup> = 1.<\/p>\n

Question 42.
\nFind the coefficient of x6<\/sup> in (3 + 2x + x2<\/sup>)6<\/sup>.
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 43.
\nIf n is a positive integer, then prove that
\nCo<\/sub> + \\(\\frac{C_{1}}{2}\\) + \\(\\frac{C_{2}}{3}\\) + ….. + \\(\\frac{C_{n}}{n+1}\\) = \\(\\frac{2^{n+1}-1}{n+1}\\)
\nSolution:
\n\"Inter<\/p>\n

Question 44.
\nIf n is a positive integer and x is any nonzero real number, then prove that
\n\"Inter
\n(May. ’14, May 13, ’05)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 45.
\nProve that
\n\"Inter
\nSolution:
\n\"Inter
\nNow we can find the term independent of in the L.H.S. of (1).
\n\"Inter
\nSuppose n is an even integer, say n = 2k. Then from (2),
\n\"Inter
\nWhen n is odd:
\nObserve that the expansion in the numerator of (2) contains only even powers of x.
\n\u2234 If n is odd, then there is no constant term in (2) (i.e.,) the term indep. of x in
\n\"Inter<\/p>\n

Question 46.
\nFind the set E of the value of x for which the binomial expansions for the following are valid
\n(i) (3 – 4x)3\/4<\/sup>
\n(ii) (2 + 5)x-1\/2<\/sup>
\n(iii) (7 – 4x)-5<\/sup>
\n(iv) (4 + 9x)-2\/3<\/sup>
\n(iv) (a + bx)r<\/sup> (Mar. ’08)
\nSolution:
\ni) (3 – 4x)3\/4<\/sup> = 33\/4<\/sup>\\(\\left(1-\\frac{4 x}{3}\\right)^{3 \/ 4}\\)
\nThe binomial expansion of (3 – 4x)3\/4<\/sup> is valid, when \\(\\frac{4 x}{3}\\) < 1
\ni.e., |x| < \\(\\frac{3}{4}\\)
\ni.e., E = \\(\\left(\\frac{-3}{4}, \\frac{3}{4}\\right)\\)<\/p>\n

ii) (2 + 5x)-1\/2<\/sup> = 2-1\/2<\/sup>\\(\\left(1+\\frac{5 x}{2}\\right)^{-1 \/ 2}\\)
\nThe binomial expansion of (2 + 5x)-1\/2<\/sup> is valid when |\\(\\frac{5 x}{3}\\)| < 1 \u21d2 |x| < \\(\\frac{2}{5}\\)
\ni.e., E = (-\\(\\frac{2}{5}\\), \\(\\frac{2}{5}\\))<\/p>\n

iii) (7 – 4x)-5<\/sup> = 7-5<\/sup>\\(\\left(1-\\frac{4 x}{7}\\right)^{-5}\\)
\nThe binomial expansion of (7 – 4x)-5<\/sup> is valid when \\(\\frac{4 x}{7}\\) < 1 \u21d2 |x| < \\(\\frac{7}{4}\\)
\ni.e., E = \\(\\left(\\frac{-7}{4}, \\frac{7}{4}\\right)\\)<\/p>\n

iv) (4 + 9x)-2\/3<\/sup> = 4-2\/3<\/sup> \\(\\left(1+\\frac{9 x}{4}\\right)^{-2 \/ 3}\\)
\nThe binomial expansipn of (4 + 9x)-2\/3<\/sup> is valid
\nWhen \\(\\frac{9 x}{4}\\) < 1
\n\u21d2 |x| < \\(\\frac{4}{9}\\)
\n\u21d2 x \u2208 \\(\\left(\\frac{-4}{9}, \\frac{4}{9}\\right)\\)
\ni.e., E = \\(\\left(\\frac{-4}{9}, \\frac{4}{9}\\right)\\)<\/p>\n

v) For any non zero reals a and b, the set of x for which the binomial expansion of (a + bx)r<\/sup> is valid when r \u2209 Z+<\/sup> \u222a {0}, is \\(\\left(-\\frac{|a|}{|b|}, \\frac{|a|}{|b|}\\right)\\)<\/p>\n

\"Inter<\/p>\n

Question 47.
\nFind the
\n(i) 9th<\/sup> term of \\(\\left(2+\\frac{x}{3}\\right)^{-5}\\)
\n(ii) 10th<\/sup> term of \\(\\left(1-\\frac{3 x}{4}\\right)^{4 \/ 5}\\)
\n(iii) 8th<\/sup> term of \\(\\left(1-\\frac{5 x}{2}\\right)^{-3 \/ 5}\\)
\n(iv) 6th<\/sup> term of \\(\\left(3+\\frac{2 x}{3}\\right)^{3 \/ 2}\\)<\/p>\n

(i) 9th<\/sup> term of \\(\\left(2+\\frac{x}{3}\\right)^{-5}\\)
\nSolution:
\n\"Inter
\nwe get X = \\(\\frac{x}{6}\\), n = 5
\nThe general term in the binomial expansion of (1 + x)-n<\/sup> is
\n\"Inter<\/p>\n

ii) 10th<\/sup> term of \\(\\left(1-\\frac{3 x}{4}\\right)^{4 \/ 5}\\)
\nSolution:
\n\"Inter<\/p>\n

iii) 8th<\/sup> term of \\(\\left(1-\\frac{5 x}{2}\\right)^{-3 \/ 5}\\)
\nSolution:
\n\"Inter<\/p>\n

iv) 6th<\/sup> term of \\(\\left(3+\\frac{2 x}{3}\\right)^{3 \/ 2}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 48.
\nWrite the first 3 terms in the expansion of
\n(i) \\(\\left(1+\\frac{x}{2}\\right)^{-5}\\)
\n(ii) (3 + 4x)-2\/3<\/sup>
\n(iii) (4 – 5x)-1\/2<\/sup><\/p>\n

i) \\(\\left(1+\\frac{x}{2}\\right)^{-5}\\)
\nSolution:
\nWe have
\n\"Inter<\/p>\n

ii) (3 + 4x)-2\/3<\/sup>
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

iii) (4 – 5x)-1\/2<\/sup>
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

iv) (2 – 3x)-1\/3<\/sup>
\nSolution:
\n\"Inter<\/p>\n

Question 49.
\nFind the coefficient of x12<\/sup> in \\(\\frac{1+3 x}{(1-4 x)^{4}}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 50.
\nFind coeff. of x6<\/sup> in the expansion of (1 – 3x)-2\/5<\/sup>
\nSolution:
\n\"Inter<\/p>\n

Question 51.
\nFind the sum of the infinite series
\n\"Inter
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 52.
\nFind the sum of the series
\n\"Inter
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 53.
\nIf x = \\(\\frac{1}{5}\\) + \\(\\frac{1.3}{5.10}\\) + \\(\\frac{1.3 .5}{5.10 .15}\\) + ……. \u221e then find 3x2<\/sup> + 6x. (Mar. ’14, ’07, ’06; May. ’11)
\nSolution:
\nGiven that
\n\"Inter
\n\"Inter<\/p>\n

Question 54.
\nFind an approximate value of
\ni) \\(\\frac{1}{\\sqrt[3]{999}}\\)
\nii) (627)1\/4<\/sup>
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 55.
\nIf |x| is so small that x3 and higher powers or x can be neglected, find approximate value of \\(\\frac{(4-7 x)^{1 \/ 2}}{(3+5 x)^{3}}\\)
\nSolution:
\n\"Inter<\/p>\n

Question 56.
\nFind an approximate value of \\(\\sqrt[6]{63}\\) correct to 4 decimal places.
\nSolution:
\n\"Inter<\/p>\n

Question 57.
\nIf |x| is so small th\u00e1t x2<\/sup> and higher powers of x may be neglected, then find an approximate value of
\n\"Inter
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 58.
\nIf |x| is so small that x4<\/sup> and higher powers of x may be neglected, then find the approximate value of
\n\\(\\sqrt[4]{x^{2}+81}\\) – \\(\\sqrt[4]{x^{2}+16}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 59.
\nSuppose that x and y are positive and x is very small when compared to y. Then find an approximate value of
\n\"Inter
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 60.
\nExpand \\(5 \\sqrt{5}\\) in increasing powers of \\(\\frac{4}{5}\\)
\nSolution:
\n\"Inter<\/p>\n","protected":false},"excerpt":{"rendered":"

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Binomial Theorem Important Questions which are most likely to be asked in the exam. Intermediate 2nd Year Maths 2A Binomial Theorem Important Questions Question 1. Find the number of terms in the expression of (2x + 3y + z)7 [Mar. 14, 13, 07] … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22097"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=22097"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22097\/revisions"}],"predecessor-version":[{"id":22098,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22097\/revisions\/22098"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=22097"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=22097"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=22097"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}