Students must practice these TS Inter 2nd Year Maths 2A Important Questions Chapter 6 Binomial Theorem to help strengthen their preparations for exams.

## TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions

Question 1.

Find the largest binomial coefficient(s) in the expansion of i) (1+x)^{19} (ii) (1 +x)^{24}

Solution:

i) Here n = 19, an odd integer. Therefore, by corollary 6.1.19. the largest binomial coefficients are \({ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{\mathrm{n}-1}{2}\right)}\) and

ii) Here n 24 ¡s an even integer. Hence there is only one largest binomial coefficient, that is \({ }^n C_{\left(\frac{n}{2}\right)}={ }^{24} C_{12}\)

Question 2.

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

Solution:

Here n = 22 is an even integer. Therefore, there is only one largest binomial coefficient

Question 3.

Find the 7th term in the expansion of

\(\left(\frac{4}{x^3}+\frac{x^2}{2}\right)^{14}\)

Solution:

The general term in the expansion of (X + a)^{n} is given by

Question 4.

Find the 3rd term from the end in the expension of \(\left(x^{\frac{-2}{3}}-\frac{3}{x^2}\right)^8\)

Solution:

Comparing the given expansion with (x + a), we get

\(X=x^{\frac{-2}{3}}, a=\frac{-3}{x^2}, n=8\)

The expansion has (n + 1) = 9 terms.

Hence the 3 term from the end is 7th term from the beginning and

Question 5.

Find the coefficients of x^{9} and x^{10} in the expansion of \(\left(2 x^2-\frac{1}{x}\right)^{20}\)

Solution:

To find the coefficient of x^{9} put 40 – 3r = 9.

Then we get r = \(\frac{31}{3}\)

Since r is a positive integer this is not possible. This means that the expansion of \(\left(2 x^2-\frac{1}{x}\right)^{20}\) doesn’t posess x^{9} term. This means that the coefficient of x^{9} in the expansion of \(\left(2 x^2-\frac{1}{x}\right)^{20}\) is 0.

To find the coefficient of x^{10} put 40 – 3r = 10.

We get r = 10

Now, on substituting r 10 in (1), we get that

\(T_{11}=(-1)^{10} \cdot{ }^{20} \mathrm{C}_{10} \cdot 2^{10} \cdot x^{10}\)

Hence, the coefficient of x^{10} in the expansion of \(\left(2 x^2-\frac{1}{x}\right)^{20} \text { is }{ }^{20} \mathrm{C}_{10} \cdot 2^{10}\)

Question 6.

Find the term independent of x (that is the constant term) in the expansion of \(\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^2}\right)^{10}\)

Solution:

Question 7.

If the coefficient of x^{10 }in the expansion \(\left(a x^2+\frac{1}{b x}\right)^{11}\) is equal to the coefficient of

x^{-10} In the expansion of \(\left(a x-\frac{1}{b x^2}\right)^{11}\) find the ration between a and b where a and b are real numbers.

Solution:

Step – 1: The general term in the expansion

To find the coefficient of x^{10} in this expansion, we should consider 22 – 3r = 10 or r = 4. Hence, the coefficient of x^{10} in the expansion of

To find the coefficient of x^{-10} in this expansion

we should consider 11 – 3r = – 10 or r = 7.

Thus the coefficient of x^{-10 }in the expansion

Question 8.

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 \(\left(x^2-\frac{1}{2 x}\right)^{20}\) is given by

Question 9.

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

Solution:

The r^{th} term in the given expansion of

\((1+x)^{18} \text { is } \mathrm{T}_{\mathrm{r}}={ }^{18} \mathrm{C}_{(\mathrm{r}-1)} \cdot \mathrm{x}^{\mathrm{r}-1}\)

Thus, the coefficient of \({ }^{18} C_{r-1}\)

Given that the coefficient of (2r + 4)^{th} term = the coefficient of (r -2)^{nd} term.

That is \({ }^{18} C_{2 r+3}={ }^{18} C_{r-3}\)

⇒ 2r+ 3r – 3 or (2r + 3) +(r-3) 18

⇒ r = – 6 or r 6

Since r is a positive integer, we get r z 6

Question 10.

Prove that 2.C_{0} +7.C_{1} + 12C_{2} + …. + (5n + Z)C_{n} (5n + 4)2^{n-1}

Solution:

First method:

The coefficieint of C_{0}, C_{1}, C_{2}, ………………. C_{n} in LH.S.

are 2, 7, 12 , (5n + 2) which are in A.P. with first term a = 2 and common difference d=5

Hence from example 6.1.14 (1), we get that

2C_{0} + 7C_{1} + 12C_{2} + (5n+2).C_{n}

=(2a + nd) . 2^{n-1}

= (4 + 5n) 2^{n-1}

Second method:

The general term ((r +1)^{th} term) in LH.S (5r + 2) C_{r} Therefore,

Question 11.

Prove that

Solution:

Question 12.

For r = 0, 1, 2 ……………….. n, prove that

Solution:

Question 13.

Prove that

Solution:

Question 14.

Find the numerically greatest term(s) in the expansion of

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

(ii) (3 x-4y)^{14} when x=8, y=3

Solution:

Therefore the numerically greatest terms in the expansion of (3x – 4y)^{14} are T_{5} and T_{6}. They are

Question 15.

Prove that 6^{2n} – 35n -1 is divisible by 1225 for all natural numbers n.

Solution:

i.e., 6^{2n}-35^{n }– 1 = 1225 (k) for sorne integer k(if n≥2)

If n = 1, then 6^{2n}-35^{n }– 1 = 6^{2}-35-1 = 0, which is trivially divisible by 1225. Hence, for all natural numbers n, 6^{2}I – 35n – 1 is divisible by 1225.

Note: The above problem can also be proved by induction.

Question 16.

Suppose that n is a natural number and I, F are respectively the Integral part and fractional part of \((7+4 \sqrt{3})^n\). Then show that

(i) I Is an odd integer

(ii) (1 + F)(1 – F) = 1.

Solution:

2k where k is a positive integer ………………. (1)

Thus, I + F + f is an even integer.

Since I is an integer, we get that

F + f is an integer. Also, since 0 < F < 1 and 0 < f < 1.

we get 0 < F + f < 2.

Since F + f is an integer, we get

F + f = 1 i.e., 1 – F = f

(i) From (1), I + F+ f= 2k

⇒ 1+ 1 = 2k ⇒ 1= 2k – I, an odd integer.

(ii) (1 +F) (1-F) (1 + F) f from (2)

= \((7+4 \sqrt{3})^n(7-4 \sqrt{3})^n=(49-48)^n=1\)

Question 17.

Find the coefficient of x^{6} in (3 + 2x + x^{2})^{6
}Solution:

Question 18.

If n is a positive integer, then prove that

\(C_0+\frac{C_1}{2}+\frac{C_2}{3}+\ldots+\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1}\)

Solution:

Question 19.

If n is a positive Integer and x is any non zero real number, then prove that

Solution:

Question 20.

Prove that

Solution:

Now, we calculate the term independent of x in the L.H.S of equation (1). From (1)

Observe that the expansion in the numerator of (2) contains only even powers of x. Therefore, if n is odd, then there is no constant term in (2). In other words, the term independent of x in \((1-x)^n\left(1+\frac{1}{x}\right)^n\) is zero. Now, suppose n is an even integer say n=2 k. Then, from (2) we get

To get the term independent of x in (3), put 2 r-2 k=0. Then r=k and hence the term independent of x in

\((1-x)^n\left(1+\frac{1}{x}\right)^n\) is

\({ }^{2 \mathrm{k}} C_k \cdot(-1)^{\mathrm{k}}={ }^n C_{\frac{n}{2}}(-1)^{\frac{\mathrm{n}}{2}}\)

Question 21.

Find the set E of the values of x for which the binomial expansions for the following are valid.

(i) \((3-4 x)^{\frac{3}{4}}\)

(ii) \((2+5 x)^{\frac{-1}{2}}\)

(iii) (7 – 4x)^{-5}

(iv) \((4+9 x)^{\frac{-2}{3}}\)

(v) (a+bx)

Solution:

Question 22.

Find the

Solution:

Question 23.

Write the first 3 terms in the expansion of

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

(ii) \((3+4 x)^{\frac{-2}{3}}\)

(iii) \((4-5 x)^{\frac{-1}{2}}\)

Solution:

Question 24.

Write the general term in the expansion of

(i) \(\left(3+\frac{x}{2}\right)^{\frac{-1}{3}}\)

(ii) \(\left(2+\frac{3 x}{4}\right)^{\frac{4}{5}}\)

(iii) (1 – 4x)^{-3}

(iv) \((2-3 x)^{\frac{-1}{3}}\)

Solution:

Question 25.

Find the coefficient of x^{12} in \(\frac{(1+3 x)}{(1-4 x)^4}\)

Solution:

Question 26.

Find the coefficient of x^{6} in the expansion of \((1-3x)^{\frac{-2}{5}}\)

Solution:

Question 27.

Find the sum of the infinite series

\(1+\frac{2}{3} \cdot \frac{1}{2}+\frac{2 \cdot 5}{3 \cdot 6}\left(\frac{1}{2}\right)^2+\frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9}\left(\frac{1}{2}\right)^3+\ldots \infty\)

Solution:

Question 28.

Find the sum of the series

\(\frac{3 \cdot 5}{5 \cdot 10}+\frac{3 \cdot 5 \cdot 7}{5 \cdot 10 \cdot 15}+\frac{3 \cdot 5 \cdot 7 \cdot 9}{5 \cdot 10 \cdot 15 \cdot 20}+\ldots ……….. \infty\)

Solution:

Question 29.

If x = \(\frac{1}{5}+\frac{1 \cdot 3}{5 \cdot 10}+\frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15}+\ldots \ldots \infty\)

Solution:

Question 30.

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

Solution:

Question 31.

If \(|\mathbf{x}|\) is so small that x^{2} and higher powers of x may be neglected, then find an approximate value of \(\frac{\left(1+\frac{3 x}{2}\right)^{-4}(8+9 x)^{\frac{1}{3}}}{(1+2 x)^2}\)

Solution:

Question 32.

If |x| is so small that x^{4} and higher powers of x may be neglected, then find an approximate value of

\(\sqrt[4]{x^2+81}-\sqrt[4]{x^2+16}\)

Solution:

Question 33.

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

\(\left(\frac{y}{y+x}\right)^{\frac{3}{4}}-\left(\frac{y}{y+x}\right)^{\frac{4}{5}}\)

Solution:

Question 34

Expand \(5 \sqrt{5}\) increasing powers of \(\frac{4}{5}\)

Solution: