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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.

$${let}\:\:{S}\:\left({x}\right)={x}\:+\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{3}{x}^{\mathrm{3}} \:+...{nx}^{{n}} \:\Rightarrow\:\frac{{S}\left({x}\right)}{{x}}=\mathrm{1}+\mathrm{2}{x}\:+\mathrm{3}{x}^{\mathrm{2}} \:+....{nx}^{{n}−\mathrm{1}} \:{but} \\ $$$$\mathrm{1}+{x}\:+{x}^{\mathrm{2}} \:+...+{x}^{{n}} \:=\frac{{x}^{{n}+\mathrm{1}} −\mathrm{1}}{{x}−\mathrm{1}}\:{if}\:{x}\neq\mathrm{1}\:{let}\:{derivate}\:\Rightarrow \\ $$$$\mathrm{1}+\mathrm{2}{x}\:+\mathrm{3}{x}^{\mathrm{2}} \:=\frac{{d}}{{dx}}\left\{\:\frac{{x}^{{n}+\mathrm{1}} −\mathrm{1}}{{x}−\mathrm{1}}\right\}\:=\frac{{nx}^{{n}+\mathrm{1}} \:−\left({n}+\mathrm{1}\right){x}^{{n}} \:+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${S}\left({x}\right)\:=\:\frac{{x}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\left\{\:{n}\:{x}^{{n}+\mathrm{1}} \:−\left({n}+\mathrm{1}\right){x}^{{n}} \:+\mathrm{1}\right\} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:{n}=\mathrm{14}\:\Rightarrow{S}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\:\frac{\mathrm{1}}{\mathrm{2}.\frac{\mathrm{1}}{\mathrm{4}}}\left\{\:\mathrm{14}\:×\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{15}} }\:−\mathrm{15}\:×\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{14}} }\:+\mathrm{1}\right\} \\ $$$$=\mathrm{2}\left\{\:\frac{\mathrm{14}}{\mathrm{2}^{\mathrm{15}} }\:−\frac{\mathrm{15}}{\mathrm{2}^{\mathrm{14}} }\:+\mathrm{1}\right\}\:=\frac{\mathrm{14}}{\mathrm{2}^{\mathrm{14}} }\:−\frac{\mathrm{15}}{\mathrm{2}^{\mathrm{13}} }\:+\mathrm{2}\:\:<\mathrm{2}\:{also}\:{we}\:{can}\:{provethat}\:{S}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)>\mathrm{1},\mathrm{999}. \\ $$

Commented by peter frank last updated on 10/Sep/18

thanks

$${thanks}\: \\ $$$$ \\ $$

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

$${s}={x}+\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{4}} +....+{nx}^{{n}} \\ $$$${sx}=\:\:\:\:\:\:\:{x}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{4}} +...+\left({n}−\mathrm{1}\right){x}^{{n}} +{nx}^{{n}+\mathrm{1}} \: \\ $$$${substructing} \\ $$$${s}\left(\mathrm{1}−{x}\right)={x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...+{x}^{{n}} −{nx}^{{n}+\mathrm{1}} \\ $$$${s}\left(\mathrm{1}−{x}\right)={x}\left(\frac{\mathrm{1}−{x}^{{n}} }{\mathrm{1}−{x}}\right)−{nx}^{{n}+\mathrm{1}} \\ $$$${s}={x}\left\{\frac{\mathrm{1}−{x}^{{n}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\right\}−\frac{{nx}^{{n}+\mathrm{1}} }{\mathrm{1}−{x}} \\ $$$${when}\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${s}\rightarrow\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\right\}\:_{} {at}\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${s}\rightarrow\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4} \\ $$$${when}\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:{s}\rightarrow\mathrm{2} \\ $$

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