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Question Number 43350 by peter frank last updated on 10/Sep/18

Commented by maxmathsup by imad last updated on 10/Sep/18
$let S (x)=x +2x^2 +3x^3 +...nx^n ⇒ ((S(x))/x)=1+2x +3x^2 +....nx^(n−1) but 1+x +x^2 +...+x^n =((x^(n+1) −1)/(x−1)) if x≠1 let derivate ⇒ 1+2x +3x^2 =(d/dx){ ((x^(n+1) −1)/(x−1))} =((nx^(n+1) −(n+1)x^n +1)/((x−1)^2 )) ⇒ S(x) = (x/((x−1)^2 )){ n x^(n+1) −(n+1)x^n +1} x=(1/2) and n=14 ⇒S((1/2))= (1/(2.(1/4))){ 14 ×(1/2^(15) ) −15 ×(1/2^(14) ) +1} =2{ ((14)/2^(15) ) −((15)/2^(14) ) +1} =((14)/2^(14) ) −((15)/2^(13) ) +2 <2 also we can provethat S((1/2))>1,999.$
$l e t S (x) = x + 2 x^{2} + 3 x^{3} + . . . {n x}^{n} \Rightarrow \frac{S (x)}{x} = 1 + 2 x + 3 x^{2} + . . . . {n x}^{n - 1} b u t$ $1 + x + x^{2} + . . . + x^{n} = \frac{x^{n + 1} - 1}{x - 1} i f x \neq 1 l e t d e r i v a t e \Rightarrow$ $1 + 2 x + 3 x^{2} = \frac{d}{d x} {\frac{x^{n + 1} - 1}{x - 1}} = \frac{{n x}^{n + 1} - (n + 1) x^{n} + 1}{{(x - 1)}^{2}} \Rightarrow$ $S (x) = \frac{x}{{(x - 1)}^{2}} {n x^{n + 1} - (n + 1) x^{n} + 1}$ $x = \frac{1}{2} a n d n = 14 \Rightarrow S (\frac{1}{2}) = \frac{1}{2 . \frac{1}{4}} {14 \times \frac{1}{2^{15}} - 15 \times \frac{1}{2^{14}} + 1}$ $= 2 {\frac{14}{2^{15}} - \frac{15}{2^{14}} + 1} = \frac{14}{2^{14}} - \frac{15}{2^{13}} + 2 < 2 a l s o w e c a n p r o v e t h a t S (\frac{1}{2}) > 1, 999 .$
Commented by peter frank last updated on 10/Sep/18

$t h a n k s$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18
	$s=x+2x^2 +3x^3 +4x^4 +....+nx^n sx= x^2 +2x^3 +3x^4 +...+(n−1)x^n +nx^(n+1) substructing s(1−x)=x+x^2 +x^3 +...+x^n −nx^(n+1) s(1−x)=x(((1−x^n )/(1−x)))−nx^(n+1) s=x{((1−x^n )/((1−x)^2 ))}−((nx^(n+1) )/(1−x)) when x=(1/2) s→(1/2){(1/((1−(1/2))^2 ))} _ at x=(1/2) s→(1/2)×4 when x=(1/2) s→2$
	$s = x + 2 x^{2} + 3 x^{3} + 4 x^{4} + . . . . + {n x}^{n}$ $s x = x^{2} + 2 x^{3} + 3 x^{4} + . . . + (n - 1) x^{n} + {n x}^{n + 1}$ $s u b s t r u c t i n g$ $s (1 - x) = x + x^{2} + x^{3} + . . . + x^{n} - {n x}^{n + 1}$ $s (1 - x) = x (\frac{1 - x^{n}}{1 - x}) - {n x}^{n + 1}$ $s = x {\frac{1 - x^{n}}{{(1 - x)}^{2}}} - \frac{{n x}^{n + 1}}{1 - x}$ $w h e n x = \frac{1}{2}$ $s \to \frac{1}{2} {\frac{1}{{(1 - \frac{1}{2})}^{2}}}_{} a t x = \frac{1}{2}$ $s \to \frac{1}{2} \times 4$ $w h e n x = \frac{1}{2} s \to 2$
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