Question-15916

Question Number 15916 by tawa tawa last updated on 15/Jun/17

Commented by tawa tawa last updated on 15/Jun/17

6 a and b

Commented by tawa tawa last updated on 15/Jun/17

please help .

Answered by RasheedSoomro last updated on 16/Jun/17

6 (b) f (x) = \frac{x + 1}{(2 x - 1) (2 x + 1) (2 x + 3)}

show that

16 f (x) = \frac{3}{2 x - 1} - \frac{2}{2 x + 1} - \frac{1}{2 x + 3}

Hence or otherwise show tbat the sum of the first

n terms of the series

\frac{2}{1 \times 3 \times 5} + \frac{3}{3 \times 5 \times 7} + \frac{4}{5 \times 7 \times 9} + \dots .

is

\frac{5}{24} - \frac{4 n + 5}{8 (2 n + 1) (2 n + 3)}

- - - - - - - - - - - - - - - - - - - - - - -

A TRY \dots

Let \frac{x + 1}{(2 x - 1) (2 x + 1) (2 x + 3)} = \frac{A}{2 x - 1} + \frac{B}{2 x + 1} + \frac{C}{2 x + 3}

A (2 x + 1) (2 x + 3) + B (2 x - 1) (2 x + 3) + C (2 x - 1) (2 x + 1)

= x + 1

^{∙} Let 2 x - 1 = 0 \Rightarrow x = 1 / 2 :

A (2 (\frac{1}{2}) + 1) (2 (\frac{1}{2}) + 3) = \frac{1}{2} + 1

8 A = \frac{3}{2} \Rightarrow A = \frac{3}{16}

^{∙} Let 2 x + 1 = 0 \Rightarrow x = - 1 / 2

B (2 (- \frac{1}{2}) - 1) (2 (- \frac{1}{2}) + 3) = - \frac{1}{2} + 1

- 4 B = \frac{1}{2} \Rightarrow B = - \frac{1}{8}

^{∙} Let x = 2 x + 3 = 0 \Rightarrow x = - 3 / 2

C (2 (- \frac{3}{2}) - 1) (2 (- \frac{3}{2}) + 1) = - \frac{3}{2} + 1

8 C = - \frac{1}{2} \Rightarrow C = - \frac{1}{16}

Hence

\frac{x + 1}{(2 x - 1) (2 x + 1) (2 x + 3)} = \frac{3}{16 (2 x - 1)} - \frac{1}{8 (2 x + 1)} - \frac{1}{16 (2 x + 3)}

16 f (x) = \frac{3}{2 x - 1} - \frac{2}{2 x + 1} - \frac{1}{2 x + 3}

Now,

\underset{x = 1}{\overset{n}{Σ}} f (x) = \underset{x = 1}{\overset{n}{Σ}} (\frac{x + 1}{(2 x - 1) (2 x + 1) (2 x + 3)})

= \frac{2}{1 \times 3 \times 5} + \frac{3}{3 \times 5 \times 7} + \frac{4}{5 \times 7 \times 9} + \dots .

= \underset{x = 1}{\overset{n}{Σ}} (\frac{3}{16 (2 x - 1)} - \frac{1}{8 (2 x + 1)} - \frac{1}{16 (2 x + 3)})

= \frac{3}{16} \underset{x = 1}{\overset{n}{Σ}} (\frac{1}{2 x - 1}) - \frac{1}{8} \underset{x = 1}{\overset{n}{Σ}} (\frac{1}{2 x + 1}) - \frac{1}{16} \underset{x = 1 1}{\overset{n}{Σ}} (\frac{1}{2 x + 3})

{Can}^{'} t continue further . I^{'} m sorry!

Commented by tawa tawa last updated on 15/Jun/17

Am with you sir . thanks for your help .

Commented by mrW1 last updated on 16/Jun/17

I^{'} ll try to continue :

= \frac{3}{16} \underset{x = 1}{\overset{n}{Σ}} (\frac{1}{2 x - 1}) - \frac{1}{8} \underset{x = 1}{\overset{n}{Σ}} (\frac{1}{2 x + 1}) - \frac{1}{16} \underset{x = 1}{\overset{n}{Σ}} (\frac{1}{2 x + 3})

= \frac{2}{16} \underset{x = 1}{\overset{n}{Σ}} (\frac{1}{2 x - 1}) + \frac{1}{16} \underset{x = 1}{\sum^{n}} (\frac{1}{2 x - 1}) - \frac{1}{8} \underset{x = 1}{\overset{n}{Σ}} (\frac{1}{2 x + 1}) - \frac{1}{16} \underset{x = 1}{\overset{n}{Σ}} (\frac{1}{2 x + 3})

= \frac{1}{8} \underset{x = 1}{\overset{n}{Σ}} (\frac{1}{2 x - 1}) + \frac{1}{16} \underset{x = 1}{\sum^{n}} (\frac{1}{2 x - 1}) - \frac{1}{8} \underset{x = 1}{\overset{n}{Σ}} (\frac{1}{2 x + 1}) - \frac{1}{16} \underset{x = 1}{\overset{n}{Σ}} (\frac{1}{2 x + 3})

= \frac{1}{8} \underset{t = 0}{\overset{n - 1}{Σ}} (\frac{1}{2 t + 1}) + \frac{1}{16} \underset{t = - 1}{\sum^{n - 2}} (\frac{1}{2 t + 3}) - \frac{1}{8} \underset{x = 1}{\overset{n}{Σ}} (\frac{1}{2 x + 1}) - \frac{1}{16} \underset{x = 1}{\overset{n}{Σ}} (\frac{1}{2 x + 3})

= \frac{1}{8} \underset{t = 1}{\overset{n}{Σ}} (\frac{1}{2 t + 1}) + \frac{1}{8} [\frac{1}{2 \times 0 + 1} - \frac{1}{2 n + 1}] + \frac{1}{16} \underset{t = 1}{\sum^{n}} (\frac{1}{2 t + 3}) + \frac{1}{16} [\frac{1}{2 (- 1) + 3} + \frac{1}{2 (0) + 3} - \frac{1}{2 (n - 1) + 3} - \frac{1}{2 (n) + 3}] - \frac{1}{8} \underset{x = 1}{\overset{n}{Σ}} (\frac{1}{2 x + 1}) - \frac{1}{16} \underset{x = 1}{\overset{n}{Σ}} (\frac{1}{2 x + 3})

= \frac{1}{8} \underset{x = 1}{\overset{n}{Σ}} (\frac{1}{2 x + 1}) + \frac{1}{8} [\frac{1}{2 \times 0 + 1} - \frac{1}{2 n + 1}] + \frac{1}{16} \underset{x = 1}{\sum^{n}} (\frac{1}{2 x + 3}) + \frac{1}{16} [\frac{1}{2 (- 1) + 3} + \frac{1}{2 (0) + 3} - \frac{1}{2 (n - 1) + 3} - \frac{1}{2 (n) + 3}] - \frac{1}{8} \underset{x = 1}{\overset{n}{Σ}} (\frac{1}{2 x + 1}) - \frac{1}{16} \underset{x = 1}{\overset{n}{Σ}} (\frac{1}{2 x + 3})

= \frac{1}{8} [\frac{1}{2 \times 0 + 1} - \frac{1}{2 n + 1}] + \frac{1}{16} [\frac{1}{2 (- 1) + 3} + \frac{1}{2 (0) + 3} - \frac{1}{2 (n - 1) + 3} - \frac{1}{2 (n) + 3}]

= \frac{1}{8} [1 - \frac{1}{2 n + 1}] + \frac{1}{16} [1 + \frac{1}{3} - \frac{1}{2 n + 1} - \frac{1}{2 n + 3}]

= \frac{1}{8} + \frac{2}{8 \times 3} - \frac{1}{16} [\frac{2}{2 n + 1} + \frac{1}{2 n + 1} + \frac{1}{2 n + 3}]

= \frac{5}{24} - \frac{1}{16} [\frac{3}{2 n + 1} + \frac{1}{2 n + 3}]

= \frac{5}{24} - \frac{4 n + 5}{8 (2 n + 1) (2 n + 3)}

Commented by tawa tawa last updated on 16/Jun/17

I really appreciate your time . God bless you sir . Rasheed

Commented by tawa tawa last updated on 16/Jun/17

Wow sir MrW 1, i really appreciate sir . God bless you sir .

Commented by tawa tawa last updated on 16/Jun/17

Sir what of the 6 a and the deduce that sum to infinity is \frac{5}{24}

Commented by mrW1 last updated on 16/Jun/17

sorry, I {don}^{'} t understand what is meant

in question 6 a .

Commented by RasheedSoomro last updated on 16/Jun/17

^{∙} Have learnt something from you mrW 1 . Thank you Sir .

^{∙} Miss tawa, your question became means to increase my

knowledge . Thank you for this .

Commented by mrW1 last updated on 16/Jun/17

Mr . Rasheed, you are welcome Sir!

Answered by tawa tawa last updated on 15/Jun/17

Please help .