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 )

Commented byDwaipayan 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