Find the sum of the infinite series 1 - 4 5 + 4 · 7 5 · 10 - 4 · 7 · 10 5 · 10 · 15 + ⋯ .

Let, S = 1 - 4 5 + 4 · 7 5 · 10 - 4 · 7 · 10 5 · 10 · 15 + ⋯ = 1 + 4 1 - 1 5 + 4 · 7 1 · 2 - 1 5 2 + 4 · 7 · 10 1 · 2 · 3 - 1 5 3 + … This is in the form of 1 - x - p q = 1 + p 1 x q + p ( p + q ) 1 · 2 x q 2 + … Here, p = 4 , q = 3 , x q = - 1 5 ⇒ x = - 3 5 ∴ S = ( 1 - x ) - p q = 1 + 3 5 - 4 3 = 8 5 - 4 3 = 5 8 4 3 = 5 4 3 8 4 3 = 5 4 3 2 4 = 625 3 16 = 5 4 3 16 ∴ The sum of the infinite series 1 - 4 5 + 4 · 7 5 · 10 - 4 · 7 · 10 5 · 10 · 15 + … is 5 4 3 16 .

Find the sum of the infinite series 3 4 + 3 · 5 4 · 8 + 3 · 5 · 7 4 · 8 · 12 + ⋯ .

Let, S = 3 4 + 3 · 5 4 · 8 + 3 · 5 · 7 4 · 8 · 12 + … … = 3 1 · 1 4 + 3 · 5 1 · 2 1 4 2 + 3 · 5 · 7 1 · 2 · 3 1 4 3 + … … Add 1 on both sides, ⇒ 1 + S = 1 + 3 1 · 1 4 + 3 · 5 1 · 2 1 4 2 + … … It is in the form 1 - x - p q = 1 + p 1 x q + p ( p + q ) 1 · 2 x q 2 + … Here, p = 3 , q = 2 , x q = 1 4 ⇒ x = 1 2 ∴ 1 + S = ( 1 - x ) - p q = 1 - 1 2 - 3 2 = 1 2 - 3 2 = 2 3 2 = 8 ∴ S = 2 2 - 1 ∴ The sum of the infinite series 3 4 + 3 · 5 4 · 8 + 3 · 5 · 7 4 · 8 · 12 + … … is 2 2 - 1

Find the sum of the infinite series 3 4 · 8 - 3 · 5 4 · 8 · 12 + 3 · 5 · 7 4 · 8 · 12 · 16 - ⋯ .

Let, S = 3 4 · 8 - 3 · 5 4 · 8 · 12 + 3 · 5 · 7 4 · 8 · 12 · 16 - … = 1 · 3 4 · 8 - 1 · 3 · 5 4 · 8 · 12 + 1 · 3 · 5 · 7 4 · 8 · 12 · 16 - … Add 1 - 1 4 on both sides, ⇒ 1 - 1 4 + S = 1 - 1 4 + 1 · 3 4 · 8 - 1 · 3 · 5 4 · 8 · 12 + ⇒ 3 4 + S = 1 - 1 1 · 1 4 + 1 · 3 1 · 2 1 4 2 - 1 · 3 · 5 1 · 2 · 3 1 4 3 + … It is in the form of 1 + x - p q = 1 - p 1 x q + ( p ) ( p + q ) 1 · 2 x q 2 - ( p ) ( p + q ) ( p + 2 q ) 1 · 2 · 3 x q 3 + … Here, p = 1 , q = 2 , x q = 1 4 ⇒ x = 1 2 ∴ 3 4 + S = ( 1 + x ) - p q = 1 + 1 2 - 1 2 = 3 2 - 1 2 = 2 3 ∴ S = 2 3 - 3 4 ∴ The sum of the infinite series 3 4 + 3 · 5 4 · 8 + 3 · 5 · 7 4 · 8 · 12 + ⋯ is 2 3 - 3 4 .

If t = 4 5 + 4 · 6 5 · 10 + 4 · 6 · 8 5 · 10 · 15 + … ∞ , then prove that 9 t = 16 .

Given, t = 4 5 + 4 · 6 5 · 10 + 4 · 6 · 8 5 · 10 · 15 + … ∞ Add 1 on both sides, ⇒ t + 1 = 1 + 4 5 + 4 · 6 5 · 10 + 4 · 6 · 8 5 · 10 · 15 + … ∞ ⇒ t + 1 = 1 + 4 1 ! 1 5 + 4 · 6 2 ! 1 5 2 + 4 · 6 · 8 3 ! 1 5 3 + … ∞ It is in the form ( 1 - x ) - p q = 1 + p 1 ! x q + p ( p + q ) 2 ! x q 2 + p ( p + 1 ) ( p + 2 q ) 3 ! x q 3 + . . . . . . . ∞ Here, p = 4 , p + q = 6 , q = 2 , x q = 1 5 ⇒ t + 1 = 1 - 2 5 - 4 2 ⇒ t + 1 = 3 5 - 2 ⇒ t + 1 = 5 3 2 ⇒ t + 1 = 25 9 ⇒ 9 t + 9 = 25 ⇒ 9 t = 16 Hence, proved.

If x = 1 · 3 3 · 6 + 1 · 3 · 5 3 · 6 · 9 + 1 · 3 · 5 · 7 3 · 6 · 9 · 12 + … , then prove that 9 x 2 + 24 x = 11 .

Given, x = 1 · 3 3 · 6 + 1 · 3 · 5 3 · 6 · 9 + 1 · 3 · 5 · 7 3 · 6 · 9 · 12 + … = 1 · 3 1 · 2 1 3 2 + 1 · 3 · 5 1 · 2 · 3 1 3 3 + … = 1 + 1 1 ! 1 3 + 1 . 3 1 . 2 1 3 2 + 1 · 3 · 5 1 . 2 · 3 1 3 3 + ⋯ - 1 + 1 3 It is in the form, 1 + p 1 ! x q + p ( p + q ) 2 ! x q 2 + ⋯ · = 1 - x - p q Here, p = 1 , q = 2 , x q = 1 3 ⇒ x = 1 - 2 3 - 1 2 - 4 3 ⇒ x = 1 3 - 1 2 - 4 3 ⇒ x = 3 - 4 3 ⇒ 3 x = 3 3 - 4 ⇒ 3 x + 4 = 3 3 ⇒ 3 x + 4 2 = 27 ⇒ 9 x 2 + 24 x + 16 = 27 ⇒ 9 x 2 + 24 x = 11 Hence, proved.

Show that 1 + x 2 + x ( x - 1 ) 2 · 4 + x ( x - 1 ) ( x - 2 ) 2 · 4 · 6 + ⋯ = 1 + x 3 + x ( x + 1 ) 3 . 6 + x ( x + 1 ) ( x + 2 ) 3 . 6 . 9 + ⋯ for any non zero rational number x .

To show, 1 + x 2 + x ( x - 1 ) 2 · 4 + x ( x - 1 ) ( x - 2 ) 2 · 4 · 6 + … = 1 + x 3 + x ( x + 1 ) 3 . 6 + x ( x + 1 ) ( x + 2 ) 3 . 6 . 9 + ⋯ LHS = 1 + x 2 + x ( x - 1 ) 2 · 4 + x ( x - 1 ) ( x - 2 ) 2 · 4 · 6 + … = 1 + x 1 2 + ( x ) ( x - 1 ) 1 · 2 1 2 2 + x ( x - 1 ) ( x - 2 ) 1 · 2 · 3 1 2 3 + … It is in the form of 1 + x n = 1 + C 1 n x + C 2 n x 2 + … … = 1 + n 1 ! x + n ( n - 1 ) 1 · 2 x 2 + … Here, x = 1 2 , n = x = 1 + 1 2 x = 3 2 x Now, RHS = 1 + x 3 + x ( x + 1 ) 3 · 6 + x ( x + 1 ) ( x + 2 ) 3 · 6 · 9 + … = 1 + x 1 1 3 + ( x ) ( x + 1 ) 1 · 2 1 3 2 + ( x ) ( x + 1 ) ( x + 2 ) 1 · 2 · 3 1 3 3 + … It is in the form of 1 - x - n = 1 + n x + n ( n + 1 ) 1 · 2 x 2 + … … Here, x = 1 3 , n = x = 1 - 1 3 - x = 2 3 - x = 3 2 x ∴ LHS = RHS Hence, proved.

E M B I B E

Intermediate Second Year Mathematics Paper 2A>Algebra>Chapter 6 - Binomial Theorem>Exercise 6(b)

Andhra Pradesh Board Solutions for Exercise 2: Exercise 6(b)

Author:Andhra Pradesh Board

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh Board Solutions for Exercise 2: Exercise 6(b)

Attempt the free practice questions from Exercise 2: Exercise 6(b) with hints and solutions to strengthen your understanding. Intermediate Second Year Mathematics Paper 2A solutions are prepared by Experienced Embibe Experts.

Questions from Andhra Pradesh Board Solutions for Exercise 2: Exercise 6(b) with Hints & Solutions

EASY

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 6 - Binomial Theorem>Exercise 6(b)>Q III

Find the sum of the infinite series $1 - \frac{4}{5} + \frac{4 \cdot 7}{5 \cdot 10} - \frac{4 \cdot 7 \cdot 10}{5 \cdot 10 \cdot 15} + \dots$ .

MEDIUM

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 6 - Binomial Theorem>Exercise 6(b)>Q III

Find the sum of the infinite series $\frac{3}{4} + \frac{3 \cdot 5}{4 \cdot 8} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12} + \dots$ .

MEDIUM

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 6 - Binomial Theorem>Exercise 6(b)>Q III

Find the sum of the infinite series $\frac{3}{4 \cdot 8} - \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16} - \dots$ .

EASY

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 6 - Binomial Theorem>Exercise 6(b)>Q III

If $t = \frac{4}{5} + \frac{4 \cdot 6}{5 \cdot 10} + \frac{4 \cdot 6 \cdot 8}{5 \cdot 10 \cdot 15} + \dots \infty,$ then prove that $9 t = 16$ .

HARD

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 6 - Binomial Theorem>Exercise 6(b)>Q III

If $x = \frac{1 \cdot 3}{3 \cdot 6} + \frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9} + \frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12} + \dots,$ then prove that $9 x^{2} + 24 x = 11$ .

EASY

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 6 - Binomial Theorem>Exercise 6(b)>Q III

If $x = \frac{5}{(2!) \cdot 3} + \frac{5 \cdot 7}{(3!) \cdot 3^{2}} + \frac{5 \cdot 7 \cdot 9}{(4!) . 3^{3}} + \dots,$ then find the value of $x^{2} + 4 x$ .

EASY

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 6 - Binomial Theorem>Exercise 6(b)>Q III

The sum to infinite terms of the series $\frac{7}{5} (1 + \frac{1}{10^{2}} + \frac{1 \cdot 3}{1 \cdot 2} \cdot \frac{1}{10^{4}} + \frac{1 \cdot 3 \cdot 5}{1 \cdot 2 \cdot 3} \cdot \frac{1}{10^{6}} + \dots)$ is $\sqrt{k}$ then find the value of $k$ .

MEDIUM

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 6 - Binomial Theorem>Exercise 6(b)>Q III

Show that $1 + \frac{x}{2} + \frac{x (x - 1)}{2 \cdot 4} + \frac{x (x - 1) (x - 2)}{2 \cdot 4 \cdot 6} + \dots = 1 + \frac{x}{3} + \frac{x (x + 1)}{3.6} + \frac{x (x + 1) (x + 2)}{3.6.9} + \dots$ for any non zero rational number $x$ .

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