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Question Number 66546 by Sayantan chakraborty last updated on 17/Aug/19

Commented by Sayantan chakraborty last updated on 17/Aug/19

$urgent need$
Commented by ~ À ® @ 237 ~ last updated on 17/Aug/19

$t h a t s u m i s S_{n} = \underset{k = 1}{\sum^{n}} \frac{2^{k} k}{(k + 1) (k + 2)}$ $a s \frac{n}{(n + 1) (n + 2)} = \frac{2 (n + 1) - (n + 2)}{(n + 1) (n + 2)} = \frac{2}{n + 2} - \frac{1}{n + 1}$ $n o w S_{n} = \underset{k = 1}{\sum^{n}} (\frac{2^{k + 1}}{k + 2} - \frac{2^{k}}{k + 1}) = \underset{k = 2}{\sum^{n + 1}} \frac{2^{k}}{k + 1} - \underset{k = 1}{\sum^{n}} \frac{2^{k}}{k + 1} = \frac{2^{n + 1}}{n + 2} - 1$
Commented by Cmr 237 last updated on 17/Aug/19

$p l e a s e i n e e d e x p l i c a t i o n a t t h e e n d o f y o u r d e m o n s t r a t i o n p l e a s e$
Commented by Sayantan chakraborty last updated on 17/Aug/19

$what are you want to mean$
Commented by Cmr 237 last updated on 17/Aug/19

$t h a n k i u n d e r s t h a n k$
Commented by mathmax by abdo last updated on 17/Aug/19
$let S_n =(2/(2.3)) +((2^2 .2)/(3.4)) +((2^3 .3)/(4.5)) +....(nterms) ⇒S_n =Σ_(k=1) ^n ((k2^k )/((k+1)(k+2))) S_n =Σ_(k=1) ^n k 2^k {(1/(k+1))−(1/(k+2))} =Σ_(k=1) ^n ((k 2^k )/(k+1))−Σ_(k=1) ^n ((k 2^k )/(k+2)) =Σ_(k=1) ^n (((k+1−1)2^k )/(k+1))−Σ_(k=1) ^n (((k+2−2)2^k )/(k+2)) =Σ_(k=1) ^n 2^k −Σ_(k=1) ^n (2^k /(k+1)) −Σ_(k=1) ^n 2^k +Σ_(k=1) ^n (2^(k+1) /(k+2)) =Σ_(k=3) ^(n+2) (2^(k−1) /k) −Σ_(k=2) ^(n+1) (2^(k−1) /k) =Σ_(k=3) ^(n+1) (2^(k−1) /k) +(2^(n+1) /(n+2))−1−Σ_(k=3) ^(n+1) (2^(k−1) /k) =(2^(n+1) /(n+2))−1 the correct answer is (c)$
$l e t S_{n} = \frac{2}{2.3} + \frac{2^{2} . 2}{3.4} + \frac{2^{3} . 3}{4.5} + . . . . (n t e r m s) \Rightarrow S_{n} = \sum_{k = 1}^{n} \frac{k 2^{k}}{(k + 1) (k + 2)}$ $S_{n} = \sum_{k = 1}^{n} k 2^{k} {\frac{1}{k + 1} - \frac{1}{k + 2}} = \sum_{k = 1}^{n} \frac{k 2^{k}}{k + 1} - \sum_{k = 1}^{n} \frac{k 2^{k}}{k + 2}$ $= \sum_{k = 1}^{n} \frac{(k + 1 - 1) 2^{k}}{k + 1} - \sum_{k = 1}^{n} \frac{(k + 2 - 2) 2^{k}}{k + 2}$ $= \sum_{k = 1}^{n} 2^{k} - \sum_{k = 1}^{n} \frac{2^{k}}{k + 1} - \sum_{k = 1}^{n} 2^{k} + \sum_{k = 1}^{n} \frac{2^{k + 1}}{k + 2}$ $= \sum_{k = 3}^{n + 2} \frac{2^{k - 1}}{k} - \sum_{k = 2}^{n + 1} \frac{2^{k - 1}}{k} = \sum_{k = 3}^{n + 1} \frac{2^{k - 1}}{k} + \frac{2^{n + 1}}{n + 2} - 1 - \sum_{k = 3}^{n + 1} \frac{2^{k - 1}}{k}$ $= \frac{2^{n + 1}}{n + 2} - 1 t h e c o r r e c t a n s w e r i s (c)$
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