Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a)

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 – x²)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 xⁿ 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 xⁿ 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) x¹¹ in \(\left(2 x^2+\frac{3}{x^3}\right)^{13}\)
Solution:

(iii) x² 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.
T₆ 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) (4x² + 5x³)¹⁷
Solution:
(4x² + 5x³)¹⁷ = [x²(4 + 5x)]¹⁷ = x³⁴(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 . C₀ + 5 . C₁ + 8 . C₂ + ……… + (3n+2) . C_n = (3n + 4) . 2^n-1
Solution:

(ii) C₀ – 4 . C₁ + 7 . C₂ – 10 . C₃ + ……… = 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) C₀ + 2 . C₁ + 4 . C₂ + 8 . C₃ + ….. + 2ⁿ . C_n = 3ⁿ
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) C₀ . C₃ + C₁ . C₄ + C₂ . C₅ + …….. + C_n-3 . C_n
Solution:
We know that
(1 + x)ⁿ = C₀ + C₁ x + C₂ x² + ……. + C_n . xⁿ ……….(1)
On replacing x by \(\frac{1}{x}\), we get

(iii) 2² . C₀ + 3² . C₁ + 4² . C₂ + ……… + (n + 2)² C_n
Solution:

(iv) 3C₀ + 6C₁ + 12C₂ + ……… + 3 . 2ⁿ . C_n
Solution:

Question 7.
Using the binomial theorem, prove that 50ⁿ – 49n – 1 is divisible by 492 for all positive integers n.
Solution:

= 49² [a positive integer]
Hence 50ⁿ – 49n – 1 is divisible by 49² for all positive integers of n.

Question 8.
Using the binomial theorem, prove that 5⁴ⁿ + 52n – 1 is divisible by 676 for all positive integers n.
Solution:

∴ 5⁴ⁿ + 52n – 1 is divisible by 676, for all positive integers n.

Question 9.
If (1 + x + x²)ⁿ = a₀ + a₁ x + a₂ x² + ……… + a_2n x²ⁿ, then prove that
(i) a₀ + a₁ + a₂ + ……… + a_2n = 3ⁿ
(ii) a₀ + a₂ + a₄ + …… + a_2n = \(\frac{3^n+1}{2}\)
(iii) a₁ + a₃ + a₅ + ……… + a_2n-1 = \(\frac{3^n-1}{2}\)
(iv) a₀ + a₃ + a₆ + a₉ + ……….. = 3^n-1
Solution:

Question 10.
If (1 + x + x² + x³)⁷ = b₀ + b₁x + b₂x² + ………. b₂₁ x²¹, then find the value of
(i) b₀ + b₂ + b₄ + …….. + b₂₀
(ii) b₁ + b₃ + b₅ + ………. + b₂₁
Solution:

Question 11.
If the coefficient of x¹¹ and x¹² 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 2⁴ = 16
The remainder when 24 is divided by 17 is 1
22013 = (24)⁵⁰³ . 2¹
∴ 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 x⁹, x¹⁰, x¹¹ in the expansion of (1 + x)ⁿ are in A.P., then prove that n² – 41n + 398 = 0.
Solution:
The coefficients of x⁹, x¹⁰, x¹¹ in (1 + x)ⁿ are

⇒ (n – 9) (n – 21) = 11(n – 19)
⇒ n² – 9n – 21n + 189 = 11n – 209
⇒ n² – 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 ⁿC_r-1, ⁿC_r, ⁿC_r+1 are three successive binomial coefficients in (1 + x)ⁿ.
Then ⁿC_r-1 = 36; ⁿC_r = 84 and ⁿC_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 n² – (4r + 1)n + 4r² – 2 = 0.
Solution:
Coefficient of Tr = ⁿC_r-1

Question 5.
Find the sum of the coefficients of x³² 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) P² – Q² = (x² – a²)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 a₁, a₂, a₃, a₄ 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 a₁, a₂, a₃, a₄ are the coefficients of 4 consecutive terms in (1 + x)ⁿ respectively.
Let a₁ = ⁿC_r-1, a₂ = ⁿC_r, a₃ = ⁿC_r+1, a₄ = ⁿC_r+2

Question 8.
Prove that (²ⁿC₀)² – (²ⁿC₁)² + (²ⁿC₂)² – (²ⁿC₃)² + ……… + (²ⁿC_2n)² = (-1)^{n 2n}C_n
Solution:

Question 9.
Prove that (C₀ + C₁)(C₁ + C₂)(C₂ + C₃) ………… (C_n-1 + C_n) = \(\frac{(n+1)^n}{n !}\) . C₀ . C₁ . C₂ ……… 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 = ²ⁿC_n xⁿ

Question 12.
If (1 + 3x – 2x²)¹⁰ = a₀ + a₁x + a₂x² + …….. + a₂₀ x²⁰ then prove that
(i) a₀ + a₁ + a₂ + ……… + a₂₀ = 2¹⁰
(ii) a₀ – a₁ + a₂ – a₃ + ……….. + a₂₀ = 4¹⁰
Solution:
(1 + 3x – 2x²)¹⁰ = a₀ + a₁x + a₂x² + ……… + a₂₀ x²⁰

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 = 4ⁿ
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.