Question-104859

Question Number 104859 by qwertyu last updated on 24/Jul/20

Answered by Dwaipayan Shikari last updated on 24/Jul/20

S_n =(1/2)+(2/2^2 )+(3/2^3 )+(4/2^4 )+....+(n/2^n ) (S_n /2)= (1/2^2 )+(2/2^3 )+....+((n−1)/2^n )+(n/2^(n+1) ) (S_n /2)=((1/2)+(1/2^2 )+(1/2^3 )+....)−(n/2^(n+1) ) S_n =2.(1/2)(((1−2^(−n) )/(1/2)))−(n/2^n ) S_n =2(1−2^(−n) )−(n/2^n )

$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{4}}{\mathrm{2}^{\mathrm{4}} }+….+\frac{{n}}{\mathrm{2}^{{n}} } \\ $$$$\frac{{S}_{{n}} }{\mathrm{2}}=\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{3}} }+….+\frac{{n}−\mathrm{1}}{\mathrm{2}^{{n}} }+\frac{{n}}{\mathrm{2}^{{n}+\mathrm{1}} } \\ $$$$\frac{{S}_{{n}} }{\mathrm{2}}=\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+….\right)−\frac{{n}}{\mathrm{2}^{{n}+\mathrm{1}} } \\ $$$${S}_{{n}} =\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}−\mathrm{2}^{−{n}} }{\frac{\mathrm{1}}{\mathrm{2}}}\right)−\frac{{n}}{\mathrm{2}^{{n}} } \\ $$$${S}_{{n}} =\mathrm{2}\left(\mathrm{1}−\mathrm{2}^{−{n}} \right)−\frac{{n}}{\mathrm{2}^{{n}} } \\ $$

Commented by Dwaipayan Shikari last updated on 24/Jul/20

$${I}\:{misunderstood}\:{the}\:{first}\:{one}.\:{It}\:{is}\:{corrected} \\ $$

Answered by abdomathmax last updated on 26/Jul/20

let f(x) =Σ_(k=0) ^n x^k ⇒f(x) =((x^(n+1) −1)/(x−1)) if x≠1 ⇒Σ_(k=1) ^n kx^(k−1) =(d/dx)(...) =((nx^(n+1) −(n+1)x^n +1)/((x−1)^2 )) ⇒Σ_(k=1) ^n k x^k =(x/((1−x)^2 ))×(nx^(n+1) −(n+1)x^n +1) x=(1/2) we get Σ_(k=1) ^n (k/2^k ) =(1/(2(1−(1/2))^2 ))×((n/2^(n+1) )−((n+1)/2^n ) +1) =((2n)/2^(n+1) ) −2((n+1)/2^n ) +2 =(n/2^n )−((n+1)/2^(n−1) ) +2

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{x}^{\mathrm{k}} \:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{x}^{\mathrm{n}+\mathrm{1}} −\mathrm{1}}{\boldsymbol{\mathrm{x}}−\mathrm{1}}\:\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{x}}\neq\mathrm{1} \\ $$$$\Rightarrow\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{kx}^{\mathrm{k}−\mathrm{1}} \:=\frac{\mathrm{d}}{\mathrm{dx}}\left(…\right)\:=\frac{\mathrm{nx}^{\mathrm{n}+\mathrm{1}} −\left(\mathrm{n}+\mathrm{1}\right)\mathrm{x}^{\mathrm{n}} \:+\mathrm{1}}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{k}\:\mathrm{x}^{\mathrm{k}} \:=\frac{\mathrm{x}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} }×\left(\mathrm{nx}^{\mathrm{n}+\mathrm{1}} −\left(\mathrm{n}+\mathrm{1}\right)\mathrm{x}^{\mathrm{n}} \:+\mathrm{1}\right) \\ $$$$\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{we}\:\mathrm{get}\: \\ $$$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{k}}{\mathrm{2}^{\mathrm{k}} }\:=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }×\left(\frac{\mathrm{n}}{\mathrm{2}^{\mathrm{n}+\mathrm{1}} }−\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }\:+\mathrm{1}\right) \\ $$$$=\frac{\mathrm{2n}}{\mathrm{2}^{\mathrm{n}+\mathrm{1}} }\:−\mathrm{2}\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }\:+\mathrm{2}\:=\frac{\mathrm{n}}{\mathrm{2}^{\mathrm{n}} }−\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}^{\mathrm{n}−\mathrm{1}} }\:+\mathrm{2} \\ $$

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