Inter 2nd Year Maths 2A Binomial Theorem Important Questions

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]
Solution:
Number of terms in (a + b + c)ⁿ are \(\frac{(n+1)(n+2)}{2}\), where n is a positive integer. Hence number of terms in (2x + 3y + z)⁷ are \(\frac{(7+1)(7+2)}{2}=\frac{8 \times 9}{2}\) = 36

Question 2.
Prove that C₀ + 2 . C₁ + 4 . C₂ + 8 . C₃ + …………… + 2ⁿ . C_n = 3ⁿ [A.P. Mar. 15; May 07]
Solution:
L.H.S. = C₀ + 2 . C₁ + 4 . C₂ + 8 . C₃ + …………… + 2ⁿ . C_n
= ₀ + C₁(2) + C₂ (2)² + C₃ (2³) + …………… + C_n . 2ⁿ
= (1 + 2)ⁿ = 3ⁿ
Note: (1 + x)ⁿ = C₀ + C₁ . x + C₂ . x² + …………… + C_n . xⁿ

Question 3.
If ²²C_r is the largest binomial coefficient in the expansion of (1 + x)²², find the value of ¹³C_r. [A.P. Mar. 15; May 07]
Solution:
Here n = 22 is an even integer. There is only one largest binomial coefficient and it is
ⁿC_(n/2) = ²²C₁₁ = ²²C_r ⇒ r = 11
∴ ¹³C_r = ¹³C₁₁ = ¹³C₂ = \(\frac{13 \times 12}{1 \times 2}\) = 78

Question 4.
Write down and simplify 6^th term in (\(\frac{2x}{3}\) + \(\frac{3y}{2}\))⁹ [May 13]
Solution:
6^th term in (\(\frac{2x}{3}\) + \(\frac{3y}{2}\))⁹
The general term in (\(\frac{2x}{3}\) + \(\frac{3y}{2}\))⁹ is
T_r+1 = ⁹C_r (\(\frac{2x}{3}\))^9-r (\(\frac{3y}{2}\))^r
Put r = 5
T₆ = ⁹C₅ (\(\frac{2x}{3}\))⁴ (\(\frac{3y}{2}\))⁵
= ⁹C₅ (\(\frac{2}{3}\))⁴ (\(\frac{3}{2}\))⁵ x⁴ y⁵
= \(\frac{9 \times 8 \times 7 \times 6}{1 \times 2 \times 3 \times 4} \frac{\left(2^{4}\right)}{3^{4}} \cdot \frac{3^{5}}{2^{5}} \cdot x^{4} y^{5}\)
= 189 x⁴y⁵

Question 5.
If the coefficients of (2r + 4)^th term and (3r + 4)^th term in the expansion of (1 + x)²¹ are equal, find r. [T.S. Mar.15]
Solution:
T_2r+4 in (1 + x)²¹ is
= ²¹C_2r+3 (x)^2r+3 ………………… (1)
T_3r+4 in (1 + x)²¹ is
= ²¹C_3r+3 . (x)^3r+3 ……………….. (2)
⇒ Coefficients are equal
⇒ ²¹C_2r+3 = ²¹C_3r+3
⇒ 21 = (2r + 3) + (3r + 3)
(or)
2r + 3 = 3r + 3
⇒ 5r = 15
⇒ r = 3 (or) r = 0 .
Hence r = 0, 3.

Question 6.
Fin 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]
Solution:
The series can be written as
S = 1 + \(\frac{1}{1}\). \(\frac{1}{3}\) + \(\frac{1.3}{3.6}\) (\(\frac{1}{3}\))² + \(\frac{1.3.5}{1.2.3}\) (\(\frac{1}{3}\))³ + ……………..
The series of the right is of the form
1 + \(\frac{p}{1}\) (\(\frac{x}{q}\)) + \(\frac{p(p+q)}{1.2}\) (\(\frac{x}{q}\))² + \(\frac{p(p+q)(p+2 q)}{1.2 .3}\) (\(\frac{x}{q}\))³ + ……………
Here p = 1, q = 2, \(\frac{x}{q}\) = \(\frac{1}{3}\) ⇒ x = \(\frac{2}{3}\)
The sum of the given series
S = (1 – x)^-p/q
= (1 – \(\frac{2}{3}\))^-1/2 = (\(\frac{1}{3}\))^-1/2 = \(\sqrt{3}\)

Question 7.
Find the set E of the value of x for which the binomial expansions for the (a + bx)^r are valid. [Mar. 08]
Solution:
(4 + 9x)^-2/3 = 4^-2/3 [1 + \(\frac{9x}{4}\)]^-2/3
The binomial expansion of (4 + 9x)^-2/3 is valid
When |\(\frac{9x}{4}\)| < 1
⇒ |x| < \(\frac{4}{9}\)
⇒ x ∈ (\(\frac{-4}{9}\), \(\frac{4}{9}\))
i.e., E = (\(\frac{-4}{9}\), \(\frac{4}{9}\))

Question 8.
If the 2^nd, 3^rd and 4^th terms in the expansion of (a + x)ⁿ are respectively 240, 720,
1080, find a, x, n. [T.S. Mar. 16]
Solution:
T₂ = 240 ⇒ ⁿC₁ a^n-1 x = 240 …………….. (1)
T₃ = 720 ⇒ ⁿC₂ a^n-2 x² = 720 …………… (2)
T₄ = 1080 ⇒ ⁿC₃ a^n-3 x³ = 1080 …………… (3)
\(\frac{(2)}{(1)} \Rightarrow \frac{{ }^{n} C_{2} a^{n-2} x^{2}}{{ }^{n} C_{1} a^{n-1} x}=\begin{aligned}
&720 \\
&240
\end{aligned}\)
⇒ \(\frac{n-1}{2} \frac{x}{a}\) = 3 ⇒ (n – 1)x = 6a …………………. (4)
\(\frac{(3)}{(2)} \Rightarrow \frac{{ }^{n} C_{3} a^{n-3} x^{3}}{{ }^{n} C_{2} a^{n-2} x^{2}}=\frac{1080}{720}\)
⇒ \(\frac{n-2}{3} \frac{x}{a}=\frac{3}{2}\)
⇒ 2(n – 2)x = 9a …………………… (5)
\(\frac{(4)}{(5)} \Rightarrow \frac{(n-1) x}{2(n-2) x}=\frac{6 a}{9 a} \Rightarrow \frac{n-1}{2 n-4}=\frac{2}{3}\)
⇒ 3n – 3 = 4n – 8
⇒ n = 5
From (4), (5 – 1) x = 6a ⇒ 4x = 6a
⇒ x = \(\frac{3}{2}\) a
Substitute x = \(\frac{3}{2}\) a, n = 5 in (1)
⁵C₁ . a⁴ . \(\frac{3}{2}\) a = 240
5 × \(\frac{3}{2}\) a⁵ = 240
a⁵ = \(\frac{480}{15}\) = 32 = 2⁵
∴ a = 2, x = \(\frac{3}{2}\) a = \(\frac{3}{2}\) (2) = 3
∴ a = 2, x = 3, n = 5.

Question 9.
If the coefficients of r^th, (r + 1)^th, and (r + 2)^nd, terms in the expansion of (1 + x)ⁿ, are in A.P. then show that n² – (4r + 1)n + 4r² – 2 = 0. [T.S. Mar. 15, 08]
Solution:
Coefficient of T_r = ⁿC_r-1
Coefficient of T_r+1 = ⁿC_r
Coefficient of T_r+2 = ⁿC_r+1
Given ⁿC_r-1, ⁿC_r, ⁿC_r+1 are in A.P.
⇒ 2 . ⁿC_r = ⁿC_r-1 + ⁿC_r+1

⇒ (2n – 3r + 2) (r + 1) = (n – r) (n – r + 1)
⇒ 2nr + 2n – 3r² – 3r + 2r + 2 = n² – 2nr + r² + n – r
⇒ n² – 4nr + 4r² – n – 2 = 0
∴ n² – (4r + 1)n + 4r² – 2 = 0

Question 10.
If 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]
Solution:

= (n + 1)² Σ r – 2(n + 1) Σ r² + Σ r³
= (n + 1)² \(\frac{(n)(n+1)}{2}\)

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

ii) (5 + x)^3/2 = [5 (1 + \(\frac{x}{5}\))]^3/2 [T.S. Mar. 17]
= 5^3/2 (1 + \(\frac{x}{5}\))]^3/2
∴ The binomial expansion of (5 + x)^3/2 is valid when \(\frac{x}{5}\) < 1
(i.e.,) |x| < 5
(i.e.,) x ∈ (-5, 5)

iii) (7 + 3x)^-5 = [7 (1 + \(\frac{3}{7}\) x)]^-5
= 7^-5 (1 + \(\frac{3}{7}\) x)]^-5
(7 + 3x)^-5 is valid when \(\frac{3x}{7}\) < 1
⇒ |x| < \(\frac{7}{3}\) ⇒ x ∈ (\(\frac{-7}{3}\), \(\frac{7}{3}\))

iv) (4 – \(\frac{x}{3}\))^-1/2 = [4(1 – \(\frac{x}{3}\))]^-1/2
(4 – \(\frac{x}{3}\))^-1/2 is valid when \(\frac{-x}{12}\) < 1
⇒ |x| < 12
⇒ x ∈ (-12, 12)

Question 12.
Find the sum of the infinite series
\(\frac{3}{4}\) + \(\frac{3.5}{4.8}\) + \(\frac{3.5 .7}{4.8 .12}\) + …… (Mar. 11)
Solution:

Question 13.
If 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 + 24x = 11 (TS Mar. ’16, AP Mar. ’17, ’15)
Solution:

⇒ 3x + 4 = 3\(\sqrt{3}\)
Squaring on both sides
(3x + 4)² = (3\(\sqrt{3}\))²
⇒ 9x² + 24x + 16 = 27
⇒ 9x² + 24x = 11

Question 14.
If 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 + 4x. (mar. 13)
Solution:

Question 15.
Find the sum of the infinite series
\(\frac{7}{5}\) (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. ‘16, May 13; Mar. ’05)
Solution:

Question 16.
For n = 0, 1, 2, 3, ….n, prove that

(TS Mar. ’15)
Solution:

Question 17.
If x = \(\frac{1}{5}\) + \(\frac{1.3}{5.10}\) + \(\frac{1.3 .5}{5.10 .15}\) + …… ∞ find 3x² + 6x. (May. ’14, ’07, ’06; May. ’11)
Solution:
Given that

⇒ 3(1 + x)² = 5
⇒ 3x² + 6x + 3= 5
⇒ 3x² + 6x = 2

Question 18.
Write the expansion or (2a + 3b)⁶.
Solution:

Question 19.
Find the 5th term in the expansion of (3x – 4y)⁷.
Solution:
T₅ = T_{4 + 1}
= ⁷C₄ (3x)^{7 – 4} (-4y)⁴
= 35.27x³. 256y⁴
= 241920 x³ y⁴

Question 20.
Find the 4^th term from the end in the expansion (2a + 5b)⁸.
Solution:
(2a + 5b)⁸ expansion contain 9 terms. The fourth term from the end is 6^th term from the beginning.
∴ ⁸C₅ (2a)^{8 – 5} (5b)⁵
= ⁸C₅ (2a)^{8 – 5} (5b)⁵
= ⁸C₅ . 2³. 5⁵ a³ b⁵

Question 21.
Find the middle term of the following expansions
(i) (3a – 5b)⁶
(ii) (2x + 3y)⁷
Solution:
i) Here n = 6 (even)
∴ \(\frac{n}{2}\) + 1 = \(\frac{6}{2}\) + 1 = 4^th term is the middle term
∴ T₄ = T_{3 + 1}
= ⁶C₃ (3a)^{6 – 3} (-5b)³,
= –⁶C₃. 3³. 5³. a³ b³

ii) Here n = 7 (odd)
\(\frac{\mathrm{n}+1}{2}\) = \(\frac{7+1}{2}\) = 4, \(\frac{\mathrm{n}+3}{2}\) = \(\frac{7+3}{2}\) = 5
∴ 4^th, 5^th terms are middle terms.
∴ T₄ = T_{3 + 1} = ⁷C₃ (2x)^{7 – 3} (3y)³ = ⁷C₃ 2⁴ 3³.x⁴.y³
T₅ = T_{4 + 1} = ⁷C₄(2x)^7-4(3y)⁴ = ⁷C₄. 2³.3⁴. x³. y⁴

Question 22.
n is a positive integer then prove that
Solution:
i) C_o + C₁ + C₂ + …….. + C_n = 2ⁿ
ii) a) C_o + C₂ + C₄ + … + C_n = 2^{n – 1} if n is even
(b) C_o + C₂ + C₄ + …. + C_{n – 1} = 2^{n – 1} if n is odd.
iii) (a) C₁ + C₃ + C₅ + …. + C_{n – 1} = 2^{n – 1} if n is even.
(b) C₁ + C₃ + C₅ + …. + C_{n – 1} = 2^{n – 1} if n is odd.
Solution:
We know (1 + x)ⁿ = ⁿC₀ + ⁿC₁ x + ⁿC₂ x²+ …… + ⁿC_n xⁿ
= C₀ + C₁x + C₂x² + …… + C_nxⁿ

Question 23.
Prove that C₀ + 3.C₁ + 5.C₂ + ……… +(2n + 1). C_n = (2n + 2). 2^{n – 1}.
Solution:
Let S = C₀ + 3.C₁ + 5.C₂ + …… + (2n + 1). C_n —— (1)
By writing the terms in (1) in the reverse older, we get

Question 24.
Find the numerically greatest term in the binomial expansion of (1 – 5x)¹² when x = \(\frac{2}{3}\).
Solution:

Question 25.
Compute numerically ireatist term (s) in the expansionly of (3x – 5y)ⁿ when x = \(\frac{3}{4}\), y = \(\frac{2}{7}\) and n = 17
Solution:
Given x = \(\frac{3}{4}\), y = \(\frac{2}{7}\) and n = 17

Question 26.
Find the largest binomial coefficients (s) in the expansion of
(i) (1 + x)¹⁹
(ii) (1 + x)²⁴
Solution:
(i) Here n = 19 is an odd integer. Hence the largest binomial coefficients are

(ii) Here n = 24 is an even integer. Hence the largest binomial coefficient is

Question 27.
If ²²C_r is the largest binomial coefficient in the expansion of (1 + x)²², find the value of ¹³C_r. (A.P. Mar’16, May ‘11)
Solution:
Here n = 22 is an even integer. There is only one largest binomial coefficient and it is

Question 28.
Find the 7^th term in the expansion of \(\left(\frac{4}{x^{3}}+\frac{x^{2}}{2}\right)^{14}\)
Solution:
The general term in the expansion of

Question 29.
Find the 3^rd term from the end in the expansion of \(\left(x^{-2 / 3}-\frac{3}{x^{2}}\right)^{8}\)
Solution:
Comparing with (X + a)ⁿ, we get
X = x^-2/3, a = \(\frac{-3}{x^{2}}\), n = 8
In the given expansion \(\left(x^{-2 / 3}-\frac{3}{x^{2}}\right)^{8}\), we have n + 1 = 8 + 1 = 9 terms
Hence the 3^rd term from the end is 7^th term from the beginning.

Question 30.
Find the coefficient of x⁹ and x¹⁰ in the expansion of \(\left(2 x^{2}-\frac{1}{x}\right)^{2 c}\)
Solution:
If we write X = 2x² and a = –\(\frac{1}{x}\), then the general term in the expansion of

Since r = \(\frac{31}{3}\) which is impossible since r must be a positive integer. Thus ‘there is no term containing x⁹ in the expansion of the given expression. In otherwords the coefficient of x⁹ is ‘0’.
Now, to find the coefficient of x¹⁰.
put 40 – 3r = 10
⇒ r = 10

Question 31.
Find the term independent of x (that is the constant term) in the expansion of

Solution:

Question 32.
If the coefficients of x¹⁰ in the expansion of \(\left(a x^{2}+\frac{1}{b x}\right)^{11}\) is equal to the coefficient of x^-10 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.
Solution:
The general term in the expansion of \(\left(a x^{2}+\frac{1}{b x}\right)^{11}\) is

To find the coefficient of x¹⁰, put
22 – 3r = 10 ⇒ 3r = 12 ⇒ r = 4
Hence the coefficient of x¹⁰ in

Given that the coefficients are equal.
Hence from (1) and (2), we get

Question 33.
If the k^th term is the middle term in the expansion of \(\left(x^{2}-\frac{1}{2 x}\right)^{20}\), find T_k and T_{k + 3}.
Solution:
The general term in the expansion of

Question 34.
If the coefficients of (2r + 4)^th and (r – 2)^nd terms in the expansion of (1 + x)¹⁸ are equal, find r.
Solution:

Question 35.
Prove that 2.C₀ + 7.C₁ + 12.C₂ + …… + (5n + 2)C_n = (5n + 4)2_{n – 1}.
Solution:
First method:
The coefficients of C₀, C₁, C₂, …., C_n are in A.P. with first term a = 2, C.d (d) = 5

Second method:
General term in LH.S.

Question 36.
Prove that

Solution:

Question 37.
For n = 0, 1, 2, 3 , n, prove that C₀. C_r + C₁. C_{r + 1} + C₂. C_{r + 2} + ……. + C_{n – r}. C_n = ²ⁿC_{n + 1}. (T.S. Mar. ’15)

Solution:
We know that

Question 38.
Prove that

Solution:

Question 39.
Find the numerically greatest term (s) in the expansion of

i) (2 + 3x)¹⁰ when x = \(\frac{11}{8}\)
Solution:

ii) (3x – 4y)¹⁴ when x = 8, y = 3.
Solution:

Question 40.
Prove that 6²ⁿ – 35n – 1 is divisible by 1225 for all natural numbers of n.
Solution:
6²ⁿ – 35n – 1 = (36)ⁿ – 35n – 1
= (35 + 1)ⁿ – 35n – 1

Hence 6²ⁿ – 35n – 1 is divisible by 1225 for all integral values of n.

Question 41.
Suppose that n is a natural number and I, F are respectively the integral part and fractional part of (7 + \(\sqrt{3}\))ⁿ. Then show that
(i) I is an odd integer
(ii) (I + F) (I – F) = 1
Solution:
Given that (7 + 4\(\sqrt{3}\))ⁿ = I + F where I is an integer and 0 < F < 1

= 2k, where k is a positive integer —— (1)
Thus I + F + f is n even integer.
Since I is an integer, we get that F + f is an integer. Also since 0 < F < 1 and 0 < f < 1
⇒ 0 < F + f < 2
∵ F + 1 is an integer
We get F + f = 1
(i.e.,) I – F = f ——— (2)

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

Question 42.
Find the coefficient of x⁶ in (3 + 2x + x²)⁶.
Solution:

Question 43.
If n is a positive integer, then prove that
C_o + \(\frac{C_{1}}{2}\) + \(\frac{C_{2}}{3}\) + ….. + \(\frac{C_{n}}{n+1}\) = \(\frac{2^{n+1}-1}{n+1}\)
Solution:

Question 44.
If n is a positive integer and x is any nonzero real number, then prove that

(May. ’14, May 13, ’05)
Solution:

Question 45.
Prove that

Solution:

Now we can find the term independent of in the L.H.S. of (1).

Suppose n is an even integer, say n = 2k. Then from (2),

When n is odd:
Observe that the expansion in the numerator of (2) contains only even powers of x.
∴ If n is odd, then there is no constant term in (2) (i.e.,) the term indep. of x in

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

ii) (2 + 5x)^-1/2 = 2^-1/2\(\left(1+\frac{5 x}{2}\right)^{-1 / 2}\)
The binomial expansion of (2 + 5x)^-1/2 is valid when |\(\frac{5 x}{3}\)| < 1 ⇒ |x| < \(\frac{2}{5}\)
i.e., E = (-\(\frac{2}{5}\), \(\frac{2}{5}\))

iii) (7 – 4x)^-5 = 7^-5\(\left(1-\frac{4 x}{7}\right)^{-5}\)
The binomial expansion of (7 – 4x)^-5 is valid when \(\frac{4 x}{7}\) < 1 ⇒ |x| < \(\frac{7}{4}\)
i.e., E = \(\left(\frac{-7}{4}, \frac{7}{4}\right)\)

iv) (4 + 9x)^-2/3 = 4^-2/3 \(\left(1+\frac{9 x}{4}\right)^{-2 / 3}\)
The binomial expansipn of (4 + 9x)^-2/3 is valid
When \(\frac{9 x}{4}\) < 1
⇒ |x| < \(\frac{4}{9}\)
⇒ x ∈ \(\left(\frac{-4}{9}, \frac{4}{9}\right)\)
i.e., E = \(\left(\frac{-4}{9}, \frac{4}{9}\right)\)

v) For any non zero reals a and b, the set of x for which the binomial expansion of (a + bx)^r is valid when r ∉ Z⁺ ∪ {0}, is \(\left(-\frac{|a|}{|b|}, \frac{|a|}{|b|}\right)\)

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

(i) 9^th term of \(\left(2+\frac{x}{3}\right)^{-5}\)
Solution:

we get X = \(\frac{x}{6}\), n = 5
The general term in the binomial expansion of (1 + x)^-n is

ii) 10^th term of \(\left(1-\frac{3 x}{4}\right)^{4 / 5}\)
Solution:

iii) 8^th term of \(\left(1-\frac{5 x}{2}\right)^{-3 / 5}\)
Solution:

iv) 6^th term of \(\left(3+\frac{2 x}{3}\right)^{3 / 2}\)
Solution:

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

i) \(\left(1+\frac{x}{2}\right)^{-5}\)
Solution:
We have

ii) (3 + 4x)^-2/3
Solution:

iii) (4 – 5x)^-1/2
Solution:

iv) (2 – 3x)^-1/3
Solution:

Question 49.
Find the coefficient of x¹² in \(\frac{1+3 x}{(1-4 x)^{4}}\)
Solution:

Question 50.
Find coeff. of x⁶ in the expansion of (1 – 3x)^-2/5
Solution:

Question 51.
Find the sum of the infinite series

Solution:

Question 52.
Find the sum of the series

Solution:

Question 53.
If x = \(\frac{1}{5}\) + \(\frac{1.3}{5.10}\) + \(\frac{1.3 .5}{5.10 .15}\) + ……. ∞ then find 3x² + 6x. (Mar. ’14, ’07, ’06; May. ’11)
Solution:
Given that

Question 54.
Find an approximate value of
i) \(\frac{1}{\sqrt[3]{999}}\)
ii) (627)^1/4
Solution:

Question 55.
If |x| is so small that x3 and higher powers or x can be neglected, find approximate value of \(\frac{(4-7 x)^{1 / 2}}{(3+5 x)^{3}}\)
Solution:

Question 56.
Find an approximate value of \(\sqrt[6]{63}\) correct to 4 decimal places.
Solution:

Question 57.
If |x| is so small thát x² and higher powers of x may be neglected, then find an approximate value of

Solution:

Question 58.
If |x| is so small that x⁴ and higher powers of x may be neglected, then find the approximate value of
\(\sqrt[4]{x^{2}+81}\) – \(\sqrt[4]{x^{2}+16}\)
Solution:

Question 59.
Suppose that x and y are positive and x is very small when compared to y. Then find an approximate value of

Solution:

Question 60.
Expand \(5 \sqrt{5}\) in increasing powers of \(\frac{4}{5}\)
Solution: