Show that 1 1 . 2 . 3 + 1 3 . 4 . 5 + 1 5 . 6 . 7 + … = 1 2 log e 2 - 1 2 .

Given 1 1 . 2 . 3 + 1 3 . 4 . 5 + 1 5 . 6 . 7 + … = 1 2 log e 2 - 1 2 . = 1 2 1 1 · 2 - 1 2 · 3 + 1 3 · 4 - 1 4 · 5 + 1 5 . 6 - 1 6 · 7 + . . . . . = 1 2 1 - 1 2 - 1 2 - 1 3 + 1 3 - 1 4 - 1 4 - 1 5 + . . . . = 1 2 1 + 2 - 1 2 + 1 3 - 1 4 + 1 5 - 1 6 + ⋯ We know that, log e ( 1 + x ) = x - x 2 2 + x 3 3 - x 4 4 + ⋯ = 1 2 [ 1 + 2 [ log 2 - 1 ] = 1 2 log 2 - 1 2 Hence showed that 1 1 . 2 . 3 + 1 3 . 4 . 5 + 1 5 . 6 . 7 + … = 1 2 log e 2 - 1 2 .

Show that 1 2 + 3 2 · 1 4 + 5 3 · 1 8 + 7 4 · 1 16 + … = 2 - log e 2 .

Given that, the series is 1 2 + 3 2 · 1 4 + 5 3 · 1 8 + 7 4 · 1 16 + … General term of the given series is T r = 1 2 + ∑ r = 2 n 2 r - 1 r · 2 r ⇒ T r = 1 2 + ∑ r = 2 n 1 2 r - 1 - 1 r · 2 r = 1 2 + 1 2 - 1 2 · 2 2 + 1 2 2 - 1 3 · 2 3 + 1 2 3 - 1 4 · 2 4 + 1 2 4 - 1 5 · 2 5 = 1 + 1 2 2 + 1 2 3 + 1 2 4 + . . . . - 1 2 · 2 2 + 1 3 · 2 3 + 1 4 · 2 4 + 1 5 · 2 5 + . . . . We know that, sum of an infinite GP is S n = a 1 - r = 3 2 - 1 2 · 2 2 + 1 3 · 2 3 + 1 4 · 2 4 + 1 5 · 2 5 + . . . . ⇒ 3 2 - 1 2 2 2 + 1 2 3 3 + 1 2 2 4 + 1 2 2 5 + . . . . = 3 2 + 1 2 - 1 2 + 1 2 2 2 + 1 2 3 3 + 1 2 2 4 + 1 2 2 5 + . . . . We know that, - log e ( 1 - x ) = x + x 2 2 + x 3 3 + x 4 4 + ⋯ ⇒ 2 - - log 1 - 1 2 ⇒ 2 + log 1 - log 2 , ∵ log 1 = 0 = 2 - log 2 = RHS Hence, proved.

Show that log e 3 = 1 + 1 3 . 2 2 + 1 5 . 2 4 + 1 7 . 2 6 + …

Given that, the series is 1 + 1 3 . 2 2 + 1 5 . 2 4 + 1 7 . 2 6 + … General term of the given series is T r = ∑ r = 0 n 1 ( 2 r + 1 ) 2 2 r We know that, log e ( 1 + x ) = x - x 2 2 + x 3 3 - x 4 4 + ⋯ and - log e ( 1 - x ) = x + x 2 2 + x 3 3 + x 4 4 + ⋯ ⇒ log e ( 1 + x ) - log e ( 1 - x ) = x - x 2 2 + x 3 3 - x 4 4 + ⋯ + x + x 2 2 + x 3 3 + x 4 4 + ⋯ Now, substitute x = 1 2 in the above expansion we get log e 1 + 1 2 - log e 1 - 1 2 = 2 1 2 + 1 2 3 3 + 1 2 5 5 + ⋯ ⇒ log e 3 2 - log e 1 2 = 1 + 1 2 2 3 + 1 2 4 5 + ⋯ ⇒ 1 + 1 3 . 2 2 + 1 5 . 2 4 + 1 7 . 2 6 + … = log e 3 Hence, proved.

If y = x + x 2 2 + x 3 3 + x 4 4 + … , then show that x = y - y 2 2 ! + y 3 3 ! - y 4 4 ! + …

Given that, y = x + x 2 2 + x 3 3 + x 4 4 + ⋯ We know that, - log e ( 1 - x ) = x + x 2 2 + x 3 3 + x 4 4 + ⋯ ⇒ - y = log e ( 1 - x ) ⇒ x = 1 - e - y We know that, e y = 1 + y 1 ! + y 2 2 ! + ⋯ · ∞ = 1 - 1 - y 1 ! + y 2 2 ! - y 3 3 ! + y 4 4 ! + · · x = y - y 2 2 ! + y 3 3 ! - y 4 4 ! + · · · · · ∞ Hence, proved.

Show that 1 + 1 3 1 2 2 + 1 5 1 2 4 + 1 7 1 2 6 + … = log e 3 .

Given that, the series is 1 + 1 3 . 2 2 + 1 5 . 2 4 + 1 7 . 2 6 + … General term of the given series is T r = ∑ r = 0 n 1 ( 2 r + 1 ) 2 2 r We know that, log e ( 1 + x ) = x - x 2 2 + x 3 3 - x 4 4 + ⋯ and - log e ( 1 - x ) = x + x 2 2 + x 3 3 + x 4 4 + ⋯ ⇒ log e ( 1 + x ) - log e ( 1 - x ) = x - x 2 2 + x 3 3 - x 4 4 + ⋯ + x + x 2 2 + x 3 3 + x 4 4 + ⋯ Now, substitute x = 1 2 in the above expansion we get log e 1 + 1 2 - log e 1 - 1 2 = 2 1 2 + 1 2 3 3 + 1 2 5 5 + ⋯ ⇒ log e 3 2 - log e 1 2 = 1 + 1 2 2 3 + 1 2 4 5 + ⋯ ⇒ 1 + 1 3 . 2 2 + 1 5 . 2 4 + 1 7 . 2 6 + … = log e 3 Hence, proved.

Find the sum of series x - 1 x + 1 + 1 2 x 2 - 1 ( x + 1 ) 2 + 1 3 x 3 - 1 ( x + 1 ) 3 + …

Given that, x - 1 x + 1 + 1 2 x 2 - 1 ( x + 1 ) 2 + 1 3 x 3 - 1 ( x + 1 ) 3 + ⋯ . Let, S = ( x - 1 ) ( x + 1 ) + 1 2 x 2 - 1 ( x + 1 ) 2 + 1 3 x 3 - 1 ( x + 1 ) 3 + ⋯ ⇒ S = x x + 1 - 1 ( x + 1 ) + 1 2 x 2 ( x + 1 ) 2 - 1 2 ( x + 1 ) 2 + 1 3 x 3 ( x + 1 ) 3 - 1 3 × 1 ( x + 1 ) 3 + · · ⇒ S = x x + 1 + 1 2 x x + 1 2 + 1 3 x x + 1 3 + · · · - 1 ( x + 1 ) + 1 2 ( x + 1 ) 2 + 1 3 ( x + 1 ) 3 · · · We know that, log ( 1 - x ) = - x - x 2 2 - x 3 3 - x 4 4 + ⋯ ⇒ - log 1 - x x + 1 - - log 1 - 1 x + 1 = - log 1 x + 1 + log x x + 1 = log ( x + 1 ) + log x - log ( x + 1 ) = log x ⇒ x - 1 x + 1 + 1 2 x 2 - 1 ( x + 1 ) 2 + 1 3 x 3 - 1 ( x + 1 ) 3 + ⋯ ⋯ = log x

E M B I B E

Intermediate Second Year Mathematics Paper 2A>Appendix>Chapter 11 - Exponential and Logarithmic Series>Exercise 1(b)

Telangana Board Solutions for Exercise 2: Exercise 1(b)

Author:Telangana Board

Telangana Board Mathematics Solutions for Exercise - Telangana Board Solutions for Exercise 2: Exercise 1(b)

Attempt the free practice questions from Exercise 2: Exercise 1(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 1(b) with Hints & Solutions

EASY

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 11 - Exponential and Logarithmic Series>Exercise 1(b)>Q 3

Show that $\frac{1}{1.2.3} + \frac{1}{3.4.5} + \frac{1}{5.6.7} + \dots = \frac{1}{2} \log_{e} 2 - \frac{1}{2}$ .

EASY

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 11 - Exponential and Logarithmic Series>Exercise 1(b)>Q 4

Show that $\frac{1}{2} + \frac{3}{2} \cdot \frac{1}{4} + \frac{5}{3} \cdot \frac{1}{8} + \frac{7}{4} \cdot \frac{1}{16} + \dots = 2 - \log_{e} 2$ .

EASY

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 11 - Exponential and Logarithmic Series>Exercise 1(b)>Q 5

Show that $\log_{e} 3 = 1 + \frac{1}{3 . 2^{2}} + \frac{1}{5 . 2^{4}} + \frac{1}{7 . 2^{6}} + \dots$

EASY

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 11 - Exponential and Logarithmic Series>Exercise 1(b)>Q 6

If $y = x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \frac{x^{4}}{4} + \dots$ , then show that $x = y - \frac{y^{2}}{2!} + \frac{y^{3}}{3!} - \frac{y^{4}}{4!} + \dots$

EASY

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 11 - Exponential and Logarithmic Series>Exercise 1(b)>Q 7

Show that $1 + \frac{1}{3} {(\frac{1}{2})}^{2} + \frac{1}{5} {(\frac{1}{2})}^{4} + \frac{1}{7} {(\frac{1}{2})}^{6} + \dots = \log_{e} 3$ .

EASY

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 11 - Exponential and Logarithmic Series>Exercise 1(b)>Q 8

Express $\log_{e} (1 + 3 x + 2 x^{2})$ in an infinite series in ascending powers of $x$ . Show that the expansion is valid only if $| x | < \frac{1}{2}$ .