Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Binomial Theorem Solutions Exercise 6(a) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2A Binomial Theorem Solutions Exercise 6(a)

I.

Question 1.

Expand the following using the binomial theorem.

(i) (4x + 5y)^{7}

Solution:

(ii) \(\left(\frac{2}{3} x+\frac{7}{4} y\right)^5\)

Solution:

(iii) \(\left(\frac{2 p}{5}-\frac{3 q}{7}\right)^6\)

Solution:

\(\sum_{r=0}^6(-1)^{r \cdot 6} C_r\left(\frac{2 p}{5}\right)^{6-r}\left(\frac{3 q}{7}\right)^r\)

(iv) (3 + x – x^{2})^{4}

Solution:

Question 2.

Write down and simplify

(i) 6th term in \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\)

Solution:

6th term in \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\)

The general term in \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\) is

(ii) 7th term in (3x – 4y)^{10}

Solution:

7th term in (3x – 4y)^{10}

The general term in (3x – 4y)^{10} is

(iii) 10th term in \(\left(\frac{3 p}{4}-5 q\right)^{14}\)

Solution:

10th term in \(\left(\frac{3 p}{4}-5 q\right)^{14}\)

General term in \(\left(\frac{3 p}{4}-5 q\right)^{14}\) is

(iv) r^{th} term in \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\) (1 ≤ r ≤ 9)

Solution:

r^{th} term in \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\)

The general term in \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\) is

Question 3.

Find the number of terms in the expansion of

(i) \(\left(\frac{3 a}{4}+\frac{b}{2}\right)^9\)

Solution:

The number of terms in (x + a)^{n} is (n + 1), where n is a positive integer.

Hence number of terms in \(\left(\frac{3 a}{4}+\frac{b}{2}\right)^9\) are 9 + 1 = 10

(ii) (3p + 4q)^{14}

Solution:

Number of terms in (3p + 4q)^{14} are 14 + 1 = 15

(iii) (2x + 3y + z)^{7}

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 4.

Find the number of terms with non-zero coefficients in (4x – 7y)^{49} + (4x + 7y)^{49}.

Solution:

∴ The number of terms with non-zero coefficient in (4x – 7y)^{49} + (4x + 7y)^{49} is 25.

Question 5.

Find the sum of the last 20 coefficients in the expansions of (1 + x)^{39}.

Solution:

∴ The sum of the last 20 coefficients in the expansion of (1 + x)^{39} is 238.

Question 6.

If A and B are coefficients of x^{n} in the expansion of (1 + x)^{2n} and (1 + x)^{2n-1} respectively, then find the value of \(\frac{A}{B}\)

Solution:

Given A and B are the coefficient of x^{n} in the expansion of (1 + x)^{2n} and (1 + x)^{2n-1} respectively.

II.

Question 1.

Find the coefficient of

(i) x^{-6} in \(\left(3 x-\frac{4}{x}\right)^{10}\)

Solution:

(ii) x^{11} in \(\left(2 x^2+\frac{3}{x^3}\right)^{13}\)

Solution:

(iii) x^{2} in \(\left(7 x^3-\frac{2}{x^2}\right)^9\)

Solution:

(iv) x^{-7} in \(\left(\frac{2 x^2}{3}-\frac{5}{4 x^5}\right)^7\)

Solution:

Question 2.

Find the term independent of x in the expansion of

(i) \(\left(\frac{\sqrt{x}}{3}-\frac{4}{x^2}\right)^{10}\)

Solution:

(ii) \(\left(\frac{3}{\sqrt[3]{x}}+5 \sqrt{x}\right)^{25}\)

Solution:

(iii) \(\left(4 x^3+\frac{7}{x^2}\right)^{14}\)

Solution:

The general term in \(\left(4 x^3+\frac{7}{x^2}\right)^{14}\) is

For term independent of x,

put 42 – 5r = 0

⇒ r = \(\frac{42}{5}\) which is not an integer.

Hence term independent of x in the given expansion is zero.

(iv) \(\left(\frac{2 x^2}{5}+\frac{15}{4 x}\right)^9\)

Solution:

Question 3.

Find the middle term(s) in the expansion of

(i) \(\left(\frac{3 x}{7}-2 y\right)^{10}\)

Solution:

The middle term in (x + a)^{n} when n is even and is \(\frac{T_{n+1}}{2}\), when n is odd, we have two middle terms, i.e., \(\frac{T_{n+1}}{2}\) and \(\frac{T_{n+3}}{2}\)

∵ n = 10 is even,

we have only one middle term (i.e.,) \(\frac{10}{2}\) + 1 = 6th term.

(ii) \(\left(4 a+\frac{3}{2} b\right)^{11}\)

Solution:

Here n = 11 is an odd integer,

we have two middle terms, i.e., \(\frac{n+1}{2}\) and \(\frac{n+3}{2}\) terms

= 6th and 7th terms are middle terms.

T_{6} in \(\left(4 a+\frac{3}{2} b\right)^{11}\) is \({ }^{11} C_5(4 a)^6\left(\frac{3}{2} b\right)^5\)

(iii) (4x^{2} + 5x^{3})^{17}

Solution:

(4x^{2} + 5x^{3})^{17} = [x^{2}(4 + 5x)]^{17} = x^{34}(4 + 5x)^{17} ……..(1)

Consider (4 + 5x)^{17}

∵ n = 17 is an odd positive integer, we have two middle terms.

They are \(\left(\frac{17+1}{2}\right)^{\text {th }}\) and \(\left(\frac{17+3}{2}\right)^{\text {th }}\) (i.e.,) 9th and 10th terms are middle terms.

(iv) \(\left(\frac{3}{a^3}+5 a^4\right)^{20}\)

Solution:

Here n = 20 is an even positive integer, we have only one middle term

Question 4.

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

(i) (4 + 3x)^{15} when x = \(\frac{7}{2}\)

Solution:

(ii) (3x + 5y)^{12} when x = \(\frac{1}{2}\), y = \(\frac{4}{3}\)

Solution:

(iii) (4a – 6b)^{13} when a = 3, b = 5

Solution:

(iv) (3 + 7x)^{n} when x = \(\frac{4}{5}\), n = 15

Solution:

Question 5.

Prove the following.

(i) 2 . C_{0} + 5 . C_{1} + 8 . C_{2} + ……… + (3n+2) . C_{n} = (3n + 4) . 2^{n-1}

Solution:

(ii) C_{0} – 4 . C_{1} + 7 . C_{2} – 10 . C_{3} + ……… = 0, if n is an even positive integer.

Solution:

(iii) \(\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\frac{C_7}{8}+\ldots \ldots=\frac{2^n-1}{n+1}\)

Solution:

(iv) \(C_0+\frac{3}{2} \cdot C_1+\frac{9}{3} \cdot C_2+\frac{27}{4} \cdot C_3\) + ……… + \(\frac{3^n}{n+1} \cdot C_n=\frac{4^{n+1}-1}{3(n+1)}\)

Solution:

(v) C_{0} + 2 . C_{1} + 4 . C_{2} + 8 . C_{3} + ….. + 2^{n} . C_{n} = 3^{n}

Solution:

Question 6.

Find the sum of the following.

(i) \(\frac{{ }^{15} C_1}{{ }^{15} C_0}+2 \frac{{ }^{15} C_2}{{ }^{15} C_1}+3 \frac{{ }^{15} C_3}{{ }^{15} C_2}\) + …….. + \(15 \frac{{ }^{15} C_{15}}{{ }^{15} C_{14}}\)

Solution:

(ii) C_{0} . C_{3} + C_{1} . C_{4} + C_{2} . C_{5} + …….. + C_{n-3} . C_{n}

Solution:

We know that

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

On replacing x by \(\frac{1}{x}\), we get

(iii) 2^{2} . C_{0} + 3^{2} . C_{1} + 4^{2} . C_{2} + ……… + (n + 2)^{2} C_{n}

Solution:

(iv) 3C_{0} + 6C_{1} + 12C_{2} + ……… + 3 . 2^{n} . C_{n}

Solution:

Question 7.

Using the binomial theorem, prove that 50^{n} – 49n – 1 is divisible by 492 for all positive integers n.

Solution:

= 49^{2} [a positive integer]

Hence 50^{n} – 49n – 1 is divisible by 49^{2} for all positive integers of n.

Question 8.

Using the binomial theorem, prove that 5^{4n} + 52n – 1 is divisible by 676 for all positive integers n.

Solution:

∴ 5^{4n} + 52n – 1 is divisible by 676, for all positive integers n.

Question 9.

If (1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + ……… + a_{2n} x^{2n}, then prove that

(i) a_{0} + a_{1} + a_{2} + ……… + a_{2n} = 3^{n}

(ii) a_{0} + a_{2} + a_{4} + …… + a_{2n} = \(\frac{3^n+1}{2}\)

(iii) a_{1} + a_{3} + a_{5} + ……… + a_{2n-1} = \(\frac{3^n-1}{2}\)

(iv) a_{0} + a_{3} + a_{6} + a_{9} + ……….. = 3^{n-1}

Solution:

Question 10.

If (1 + x + x^{2} + x^{3})^{7} = b_{0} + b_{1}x + b_{2}x^{2} + ………. b_{21} x^{21}, then find the value of

(i) b_{0} + b_{2} + b_{4} + …….. + b_{20}

(ii) b_{1} + b_{3} + b_{5} + ………. + b_{21}

Solution:

Question 11.

If the coefficient of x^{11} and x^{12} in the binomial expansion of \(\left(2+\frac{8 x}{3}\right)^n\) are equal, find n.

Solution:

The general term of \(\left(2+\frac{8 x}{3}\right)^n\) is \(T_{r+1}={ }^n C_r(2)^{n-r}\left(\frac{8 x}{3}\right)^r\)

Question 12.

Find the remainder when 22013 is divided by 17.

Solution:

We know 2^{4} = 16

The remainder when 24 is divided by 17 is 1

22013 = (24)^{503} . 2^{1}

∴ The remainder when 22013 is divided by 17 is (-1)^{503} . 2 = (-1) . 2 = -2

Question 13.

If the coefficients of (2r + 4)^{th} term and (3r + 4)^{th} term in the expansion of (1 + x)^{21} are equal, find r.

Solution:

III.

Question 1.

If the coefficients of x^{9}, x^{10}, x^{11} in the expansion of (1 + x)^{n} are in A.P., then prove that n^{2} – 41n + 398 = 0.

Solution:

The coefficients of x^{9}, x^{10}, x^{11} in (1 + x)^{n} are

⇒ (n – 9) (n – 21) = 11(n – 19)

⇒ n^{2} – 9n – 21n + 189 = 11n – 209

⇒ n^{2} – 41n + 398 = 0

Question 2.

If 36, 84, 126 are three successive binomial coefficients in the expansion of (1 + x)^{n}, find n.

Solution:

Let ^{n}C_{r-1}, ^{n}C_{r}, ^{n}C_{r+1} are three successive binomial coefficients in (1 + x)^{n}.

Then ^{n}C_{r-1} = 36; ^{n}C_{r} = 84 and ^{n}C_{r+1} = 126

Question 3.

If the 2nd, 3rd and 4th terms in the expansion of (a + x)^{n} are respectively 240, 720, 1080, find a, x, n.

Solution:

Question 4.

If the coefficients of rth, (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.

Solution:

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

Question 5.

Find the sum of the coefficients of x^{32} and x^{-18} in the expansion of \(\left(2 x^3-\frac{3}{x^2}\right)^{14}\)

Solution:

Question 6.

If P and Q are the sums of odd terms and the sum of even terms respectively in the expansion of (x + a)^{n} then prove that

(i) P^{2} – Q^{2} = (x^{2} – a^{2})^{n}

(ii) 4PQ = (x + a)^{2n} – (x – a)^{2n}

Solution:

Question 7.

If the coefficients of 4 consecutive terms in the expansion of (1 + x)^{n} are a_{1}, a_{2}, a_{3}, a_{4} respectively, then show that \(\frac{a_1}{a_1+a_2}+\frac{a_3}{a_3+a_4}=\frac{2 a_2}{a_2+a_3}\)

Solution:

Given a_{1}, a_{2}, a_{3}, a_{4} are the coefficients of 4 consecutive terms in (1 + x)^{n} respectively.

Let a_{1} = ^{n}C_{r-1}, a_{2} = ^{n}C_{r}, a_{3} = ^{n}C_{r+1}, a_{4} = ^{n}C_{r+2}

Question 8.

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

Solution:

Question 9.

Prove that (C_{0} + C_{1})(C_{1} + C_{2})(C_{2} + C_{3}) ………… (C_{n-1} + C_{n}) = \(\frac{(n+1)^n}{n !}\) . C_{0} . C_{1} . C_{2} ……… C_{n}

Solution:

Question 10.

Find the term independent of x in \((1+3 x)^n\left(1+\frac{1}{3 x}\right)^n\)

Solution:

Question 11.

Show that the middle term in the expansion of (1 + x)^{2n} is \(\frac{1.3 .5 \ldots(2 n-1)}{n !}(2 x)^n\)

Solution:

The expansion of (1 + x)^{2n} contains (2n + 1) terms.

middle term = ^{2n}C_{n} x^{n}

Question 12.

If (1 + 3x – 2x^{2})^{10} = a_{0} + a_{1}x + a_{2}x^{2} + …….. + a_{20} x^{20} then prove that

(i) a_{0} + a_{1} + a_{2} + ……… + a_{20} = 2^{10}

(ii) a_{0} – a_{1} + a_{2} – a_{3} + ……….. + a_{20} = 4^{10}

Solution:

(1 + 3x – 2x^{2})^{10} = a_{0} + a_{1}x + a_{2}x^{2} + ……… + a_{20} x^{20}

Question 13.

If (3√3 + 5)^{2n+1} = x and f = x – [x] where ([x] is the integral part of x), find the value of x.f.

Solution:

Question 14.

If R, n are positive integers, n is odd, 0 < F < 1 and if (5√5 + 11)^{n} = R + F, then prove that

(i) R is an even integer and

(ii) (R + F) . F = 4^{n}

Solution:

(i) Since R, n are positive integers, 0 < F < 1 and (5√5 + 11)^{n} = R + F

Let (5√5 – 11)^{n} = f

Now, 11 < 5√5 < 12

⇒ 0 < 5√5 – 11 < 1

⇒ 0 < (5√5 – 11)^{n} < 1

⇒ 0 < f < 1

Question 15.

If I, n are positive integers, 0 < f < 1 and if (7 + 4√3 )^{n} = I + f, then show that

(i) I is an odd integer and

(ii) (I + f) (1 – f) = 1

Solution:

Given I, n are positive integers and

(7 + 4√3)^{n} = I + f, 0 < f < 1

Let 7 – 4√3 = F

Now 6 < 4√3 < 7

⇒ -6 > -4√3 > -7

⇒ 1 > 7 – 4√3 > 0

⇒ 0 < (7 – 4√3)^{n} < 1

∴ 0 < F < 1

I + f + F = (7 + 4√3)^{n} + (7 – 4√3)^{n}

= 2k where k is an integer.

∴ I + f + F is an even integer.

⇒ f + F is an integer since I is an integer.

But 0 < f < 1 and 0 < F < 1

⇒ 0 < f + F < 2

∴ f + F = 1 ………..(1)

⇒ I + 1 is an even integer.

∴ I is an odd integer.

(I + f) (I – f) = (I + f) F …..[By (1)]

= (7 + 4√3)^{n} (7 – 4√3)^{n}

= [(7 + 4√3) (7 – 4√3)]^{n}

= (49 – 48)^{n}

= 1

Question 16.

If n is a positive integer, prove that \(\sum_{r=1}^n r^3\left(\frac{{ }^n C_r}{{ }^n C_{r-1}}\right)^2=\frac{(n)(n+1)^2(n+2)}{12}\)

Solution:

Question 17.

Find the number of irrational terms in the expansion of (5^{1/6} + 2^{1/8})^{100}.

Solution:

General term

T_{r+1} = \({ }^{100} C_r\left(5^{1 / 6}\right)^{100-r}\left(2^{1 / 8}\right)^r\) = \({ }^{100} C_r 5^{\frac{100-r}{6}} \cdot 2^{\frac{r}{8}}\)

\(\frac{100-r}{6}\) is an integer in the span

or 0 ≤ r ≤ 100 if r = 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, 70, 76, 82, 88, 94, 100

\(\frac{r}{8}\) is an integer in the span of 0 ≤ r ≤ 100

if r = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, \(\frac{100-r}{6}\), \(\frac{r}{8}\) both an integers

If r = 16, 40, 64, 88

∴ The number of rational terms in the expansion of (5^{1/6} + 2^{1/8})^{r} is 4.

∴ The number of irrational terms in the expansion of (5^{1/6} + 2^{1/8})^{r} is 101 – 4 = 97 terms.