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 C_{0} + 2 . C_{1} + 4 . C_{2} + 8 . C_{3} + …………… + 2^{n} . C_{n} = 3^{n} [A.P. Mar. 15; May 07]

Solution:

L.H.S. = C_{0} + 2 . C_{1} + 4 . C_{2} + 8 . C_{3} + …………… + 2^{n} . C_{n}

= _{0} + C_{1}(2) + C_{2} (2)^{2} + C_{3} (2^{3}) + …………… + C_{n} . 2^{n}

= (1 + 2)^{n} = 3^{n}

Note: (1 + x)^{n} = C_{0} + C_{1} . x + C_{2} . x^{2} + …………… + C_{n} . x^{n}

Question 3.

If ^{22}C_{r} is the largest binomial coefficient in the expansion of (1 + x)^{22}, find the value of ^{13}C_{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

^{n}C_{(n/2)} = ^{22}C_{11} = ^{22}C_{r} ⇒ r = 11

∴ ^{13}C_{r} = ^{13}C_{11} = ^{13}C_{2} = \(\frac{13 \times 12}{1 \times 2}\) = 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} = ^{9}C_{r} (\(\frac{2x}{3}\))^{9-r} (\(\frac{3y}{2}\))^{r}

Put r = 5

T_{6} = ^{9}C_{5} (\(\frac{2x}{3}\))^{4} (\(\frac{3y}{2}\))^{5}

= ^{9}C_{5} (\(\frac{2}{3}\))^{4} (\(\frac{3}{2}\))^{5} x^{4} y^{5}

= \(\frac{9 \times 8 \times 7 \times 6}{1 \times 2 \times 3 \times 4} \frac{\left(2^{4}\right)}{3^{4}} \cdot \frac{3^{5}}{2^{5}} \cdot x^{4} y^{5}\)

= 189 x^{4}y^{5}

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

= ^{21}C_{2r+3} (x)^{2r+3} ………………… (1)

T_{3r+4} in (1 + x)^{21} is

= ^{21}C_{3r+3} . (x)^{3r+3} ……………….. (2)

⇒ Coefficients are equal

⇒ ^{21}C_{2r+3} = ^{21}C_{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:

T_{2} = 240 ⇒ ^{n}C_{1} a^{n-1} x = 240 …………….. (1)

T_{3} = 720 ⇒ ^{n}C_{2} a^{n-2} x^{2} = 720 …………… (2)

T_{4} = 1080 ⇒ ^{n}C_{3} a^{n-3} x^{3} = 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)

^{5}C_{1} . a^{4} . \(\frac{3}{2}\) a = 240

5 × \(\frac{3}{2}\) a^{5} = 240

a^{5} = \(\frac{480}{15}\) = 32 = 2^{5}

∴ 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 n^{2} – (4r + 1)n + 4r^{2} – 2 = 0. [T.S. Mar. 15, 08]

Solution:

Coefficient of T_{r} = ^{n}C_{r-1}

Coefficient of T_{r+1} = ^{n}C_{r}

Coefficient of T_{r+2} = ^{n}C_{r+1}

Given ^{n}C_{r-1}, ^{n}C_{r}, ^{n}C_{r+1} are in A.P.

⇒ 2 . ^{n}C_{r} = ^{n}C_{r-1} + ^{n}C_{r+1}

⇒ (2n – 3r + 2) (r + 1) = (n – r) (n – r + 1)

⇒ 2nr + 2n – 3r^{2} – 3r + 2r + 2 = n^{2} – 2nr + r^{2} + n – r

⇒ n^{2} – 4nr + 4r^{2} – n – 2 = 0

∴ n^{2} – (4r + 1)n + 4r^{2} – 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 r^{2} + Σ r^{3}

= (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 9x^{2} + 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}

⇒ 9x^{2} + 24x + 16 = 27

⇒ 9x^{2} + 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 x^{2} + 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 3x^{2} + 6x. (May. ’14, ’07, ’06; May. ’11)

Solution:

Given that

⇒ 3(1 + x)^{2} = 5

⇒ 3x^{2} + 6x + 3= 5

⇒ 3x^{2} + 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:

T_{5} = T_{4 + 1}

= ^{7}C_{4} (3x)^{7 – 4} (-4y)^{4}

= 35.27x^{3}. 256y^{4}

= 241920 x^{3} y^{4}

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.

∴ ^{8}C_{5} (2a)^{8 – 5} (5b)^{5}

= ^{8}C_{5} (2a)^{8 – 5} (5b)^{5}

= ^{8}C_{5} . 2^{3}. 5^{5} a^{3} b^{5}

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

∴ T_{4} = T_{3 + 1}

= ^{6}C_{3} (3a)^{6 – 3} (-5b)^{3},

= –^{6}C_{3}. 3^{3}. 5^{3}. a^{3} b^{3}

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.

∴ T_{4} = T_{3 + 1} = ^{7}C_{3} (2x)^{7 – 3} (3y)^{3} = ^{7}C_{3} 2^{4} 3^{3}.x^{4}.y^{3}

T_{5} = T_{4 + 1} = ^{7}C_{4}(2x)^{7-4}(3y)^{4} = ^{7}C_{4}. 2^{3}.3^{4}. x^{3}. y^{4}

Question 22.

n is a positive integer then prove that

Solution:

i) C_{o} + C_{1} + C_{2} + …….. + C_{n} = 2^{n}

ii) a) C_{o} + C_{2} + C_{4} + … + C_{n} = 2^{n – 1} if n is even

(b) C_{o} + C_{2} + C_{4} + …. + C_{n – 1} = 2^{n – 1} if n is odd.

iii) (a) C_{1} + C_{3} + C_{5} + …. + C_{n – 1} = 2^{n – 1} if n is even.

(b) C_{1} + C_{3} + C_{5} + …. + C_{n – 1} = 2^{n – 1} if n is odd.

Solution:

We know (1 + x)^{n} = ^{n}C_{0} + ^{n}C_{1} x + ^{n}C_{2} x^{2}+ …… + ^{n}C_{n} x^{n}

= C_{0} + C_{1}x + C_{2}x^{2} + …… + C_{n}x^{n}

Question 23.

Prove that C_{0} + 3.C_{1} + 5.C_{2} + ……… +(2n + 1). C_{n} = (2n + 2). 2^{n – 1}.

Solution:

Let S = C_{0} + 3.C_{1} + 5.C_{2} + …… + (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 ^{22}C_{r} is the largest binomial coefficient in the expansion of (1 + x)^{22}, find the value of ^{13}C_{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 x^{9} and x^{10} in the expansion of \(\left(2 x^{2}-\frac{1}{x}\right)^{2 c}\)

Solution:

If we write X = 2x^{2} 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 x^{9} in the expansion of the given expression. In otherwords the coefficient of x^{9} is ‘0’.

Now, to find the coefficient of x^{10}.

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 x^{10} 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 x^{10}, put

22 – 3r = 10 ⇒ 3r = 12 ⇒ r = 4

Hence the coefficient of x^{10} 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.C_{0} + 7.C_{1} + 12.C_{2} + …… + (5n + 2)C_{n} = (5n + 4)2_{n – 1}.

Solution:

First method:

The coefficients of C_{0}, C_{1}, C_{2}, …., 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 C_{0}. C_{r} + C_{1}. C_{r + 1} + C_{2}. C_{r + 2} + ……. + C_{n – r}. C_{n} = ^{2n}C_{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 6^{2n} – 35n – 1 is divisible by 1225 for all natural numbers of n.

Solution:

6^{2n} – 35n – 1 = (36)^{n} – 35n – 1

= (35 + 1)^{n} – 35n – 1

Hence 6^{2n} – 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 x^{6} in (3 + 2x + x^{2})^{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 x^{12} in \(\frac{1+3 x}{(1-4 x)^{4}}\)

Solution:

Question 50.

Find coeff. of x^{6} 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 3x^{2} + 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 x^{2} and higher powers of x may be neglected, then find an approximate value of

Solution:

Question 58.

If |x| is so small that x^{4} 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: