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)

Telangana Board Solutions for Exercise 2: Exercise 6(b)

Author:Telangana Board

Telangana Board Mathematics Solutions for Exercise - Telangana 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 Telangana Board Solutions for Exercise 2: Exercise 6(b) with Hints & Solutions

EASY

12th Telangana 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 Telangana 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 Telangana 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 Telangana 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 Telangana 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 Telangana 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 Telangana 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 Telangana 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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