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Question Number 40743 by scientist last updated on 27/Jul/18

	Answered by maxmathsup by imad last updated on 27/Jul/18
	$we have Σ_(k=0) ^∞ x^k =(1/(1−x)) for ∣x∣<1 ⇒Σ_(k=1) ^∞ k x^(k−1) =(1/((1−x)^2 )) ⇒((1+x)/((1−x)^2 )) = (1+x)Σ_(k=1) ^∞ k x^(k−1) =Σ_(k=1) ^∞ k x^(k−1) +Σ_(k=1) ^∞ kx^k so the coefficient of x^(n−1) is λ =n +n−1 =2n−1 we have ((1+x)/((1−x)^2 )) = Σ_(k=0) ^∞ (k+1)x^k +Σ_(k=0) ^∞ kx^k =Σ_(k =0) ^∞ (2k+1)x^k = Σ_(k=0) ^(n−1) (2k+1)x^k +Σ_(k=n) ^∞ (2k+1)x^k ⇒ Σ_(k=n) ^∞ (2k+1)x^k =((1+x)/((1−x)^2 )) −Σ_(k=0) ^(n−1) (2k+1)x^k but Σ_(k=0) ^(n−1) (2k+1)x^k =2Σ_(k=0) ^(n−1) kx^k +Σ_(k=0) ^(n−1) x^k Σ_(k=0) ^(n−1) x^k =((1−x^n )/(1−x)) also we have Σ_(k=0) ^N x^k =((x^(N+1) −1)/(x−1)) ⇒ Σ_(k=1) ^N kx^(k−1) =((Nx^(N+1) −(N+1)x^N +1)/((1−x)^2 )) ⇒ Σ_(k=0) ^(n−1) kx^k =(x/((1−x)^2 )){(n−1)x^n −nx^(n−1) +1) ⇒ Σ_(k=n) ^∞ (2k+1)x^k =((1+x)/((1−x)^2 )) −((2x)/((1−x)^2 )){(n−1)x^n −nx^(n−1) +1}−((1−x^n )/(1−x)) ...$
	$w e h a v e \sum_{k = 0}^{\infty} x^{k} = \frac{1}{1 - x} f o r ∣ x ∣< 1 \Rightarrow \sum_{k = 1}^{\infty} k x^{k - 1} = \frac{1}{{(1 - x)}^{2}}$ $\Rightarrow \frac{1 + x}{{(1 - x)}^{2}} = (1 + x) \sum_{k = 1}^{\infty} k x^{k - 1} = \sum_{k = 1}^{\infty} k x^{k - 1} + \sum_{k = 1}^{\infty} {k x}^{k} s o t h e c o e f f i c i e n t$ $o f x^{n - 1} i s λ = n + n - 1 = 2 n - 1$ $w e h a v e \frac{1 + x}{{(1 - x)}^{2}} = \sum_{k = 0}^{\infty} (k + 1) x^{k} + \sum_{k = 0}^{\infty} {k x}^{k}$ $= \sum_{k = 0}^{\infty} (2 k + 1) x^{k} = \sum_{k = 0}^{n - 1} (2 k + 1) x^{k} + \sum_{k = n}^{\infty} (2 k + 1) x^{k} \Rightarrow$ $\sum_{k = n}^{\infty} (2 k + 1) x^{k} = \frac{1 + x}{{(1 - x)}^{2}} - \sum_{k = 0}^{n - 1} (2 k + 1) x^{k} b u t$ $\sum_{k = 0}^{n - 1} (2 k + 1) x^{k} = 2 \sum_{k = 0}^{n - 1} {k x}^{k} + \sum_{k = 0}^{n - 1} x^{k}$ $\sum_{k = 0}^{n - 1} x^{k} = \frac{1 - x^{n}}{1 - x} a l s o w e h a v e \sum_{k = 0}^{N} x^{k} = \frac{x^{N + 1} - 1}{x - 1} \Rightarrow$ $\sum_{k = 1}^{N} {k x}^{k - 1} = \frac{{N x}^{N + 1} - (N + 1) x^{N} + 1}{{(1 - x)}^{2}} \Rightarrow$ $\sum_{k = 0}^{n - 1} {k x}^{k} = \frac{x}{{(1 - x)}^{2}} {(n - 1) x^{n} - {n x}^{n - 1} + 1) \Rightarrow$ $\sum_{k = n}^{\infty} (2 k + 1) x^{k} = \frac{1 + x}{{(1 - x)}^{2}} - \frac{2 x}{{(1 - x)}^{2}} {(n - 1) x^{n} - {n x}^{n - 1} + 1} - \frac{1 - x^{n}}{1 - x} . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18

	$\frac{1 + x}{{(1 - x)}^{2}} = (1 + x) {(1 - x)}^{- 2}$ $= (1 + 2 x + 3 x^{2} + 4 x^{3} + . . . (r + 1) x^{r} + . . . .) (1 + x)$ $s o t h e t e r m s c o n t a i n i n g x^{n - 1} a r e$ $(n - 1 + 1) x^{n - 1} + x . (1 + n - 2) x^{n - 2}$ $= n . x^{n - 1} + (n - 1) x^{n - 1}$ $= (2 n - 1) x^{n - 1}$ $h e n c e t h e c o e f f i c i e n t o f x^{n - 1} i s 2 n - 1$
	Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18
	$(1+2x+3x^2 +4x^3 +...(r+1)x^r +...)(1+x) ={1+x(1+2)+x^2 (2+3)+x^3 (3+4)+..x^r (r+r+1) .... =1+3x+5x^2 +7x^3 +...(2r+1)x^r +.... so nth term is {1+(n−1)2}x^(n−1) (2n−1)x^(2n−1)$
	$(1 + 2 x + 3 x^{2} + 4 x^{3} + . . . (r + 1) x^{r} + . . .) (1 + x)$ $= {1 + x (1 + 2) + x^{2} (2 + 3) + x^{3} (3 + 4) + . . x^{r} (r + r + 1)$ $. . . .$ $= 1 + 3 x + 5 x^{2} + 7 x^{3} + . . . (2 r + 1) x^{r} + . . . .$ $s o n t h t e r m i s {1 + (n - 1) 2} x^{n - 1}$ $(2 n - 1) x^{2 n - 1}$
	Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18
	$T_(n+1) ={2n+2−1}x^(2n+2−1) =(2n+1)x^(2n+1) T_(n+2) ={2n+4−1}x^(2n+4−1) =(2n+3)x^(2n+3) T_(n+3) ={2n+6−1}x^(2n+6−1) =(2n+5)x^(2n+5) ..... .... S=T_(n+1) +T_(n+2) +T_(n+3) +.....upto ∞ =2n(x^(2n+1) +x^(2n+3) +x^(2n+5) +...)+(1.x^(2n+1) +3x^(2n+3) 5.x^(2n+5) +...) =2n.x^(2n+1) ((1/(1−x^2 )))+(1.x^(2n+1) +3.x^(2n+3) +5.x^(2n+5) ...) let S_k =1.x^(2n+1) +3.x^(2n+3) +5.x^(2n+5) +... x^2 S_k = +1.x^(2n+3) +3.x^(2n+5) +.... S_k (1−x^2 )=1.x^(2n+1) +2(x^(2n+3) +x^(2n+5) +....) S_k (1−x^2 )=1.x^(2n+1) +2.(x^(2n+3) /(1−x^2 )) S_k =(x^(2n+1) /(1−x^2 ))+((2.x^(2n+3) )/((1−x^2 )^2 )) so required sum is ((2nx^(2n+1) )/(1−x^2 ))+(x^(2n+1) /(1−x^2 ))+((2x^(2n+3) )/((1−x^2 )^2 )) =(((2n+1)x^(2n+1) )/(1−x^2 ))+((2x^(2n+3) )/((1−x^2 )^2 ))$
	$T_{n + 1} = {2 n + 2 - 1} x^{2 n + 2 - 1} = (2 n + 1) x^{2 n + 1}$ $T_{n + 2} = {2 n + 4 - 1} x^{2 n + 4 - 1} = (2 n + 3) x^{2 n + 3}$ $T_{n + 3} = {2 n + 6 - 1} x^{2 n + 6 - 1} = (2 n + 5) x^{2 n + 5}$ $. . . . .$ $. . . .$ $S = T_{n + 1} + T_{n + 2} + T_{n + 3} + . . . . . u p t o \infty$ $= 2 n (x^{2 n + 1} + x^{2 n + 3} + x^{2 n + 5} + . . .) + (1 . x^{2 n + 1} + 3 x^{2 n + 3}$ $5 . x^{2 n + 5} + . . .)$ $= 2 n . x^{2 n + 1} (\frac{1}{1 - x^{2}}) + (1 . x^{2 n + 1} + 3 . x^{2 n + 3} + 5 . x^{2 n + 5} . . .)$ $l e t$ $S_{k} = 1 . x^{2 n + 1} + 3 . x^{2 n + 3} + 5 . x^{2 n + 5} + . . .$ $x^{2} S_{k} = + 1 . x^{2 n + 3} + 3 . x^{2 n + 5} + . . . .$ $S_{k} (1 - x^{2}) = 1 . x^{2 n + 1} + 2 (x^{2 n + 3} + x^{2 n + 5} + . . . .)$ $S_{k} (1 - x^{2}) = 1 . x^{2 n + 1} + 2 . \frac{x^{2 n + 3}}{1 - x^{2}}$ $S_{k} = \frac{x^{2 n + 1}}{1 - x^{2}} + \frac{2 . x^{2 n + 3}}{{(1 - x^{2})}^{2}}$ $s o r e q u i r e d s u m i s$ $\frac{2 {n x}^{2 n + 1}}{1 - x^{2}} + \frac{x^{2 n + 1}}{1 - x^{2}} + \frac{2 x^{2 n + 3}}{{(1 - x^{2})}^{2}}$ $= \frac{(2 n + 1) x^{2 n + 1}}{1 - x^{2}} + \frac{2 x^{2 n + 3}}{{(1 - x^{2})}^{2}}$
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