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)

Andhra Pradesh Board Solutions for Exercise 2: Exercise 1(b)

Author:Andhra Pradesh Board

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh 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 Andhra Pradesh Board Solutions for Exercise 2: Exercise 1(b) with Hints & Solutions

EASY

12th Andhra Pradesh 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 Andhra Pradesh 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 Andhra Pradesh 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 Andhra Pradesh 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 Andhra Pradesh 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 Andhra Pradesh 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 $| x | < \frac{1}{2}$ .