Question Number 103560 by abony1303 last updated on 15/Jul/20

S=Σ_(k=1) ^(17) k∙2^k =?

Commented byabony1303 last updated on 15/Jul/20

pls help

Answered by mr W last updated on 15/Jul/20

S_n =Σ_(k=1) ^n k2^k   2S_n =Σ_(k=1) ^n k2^(k+1)   2S_n =Σ_(k=1) ^n (k+1)2^(k+1) −Σ_(k=1) ^n 2^(k+1)   2S_n =Σ_(k=2) ^(n+1) k2^k −2Σ_(k=1) ^n 2^k   2S_n =Σ_(k=1) ^n k2^k +(n+1)2^(n+1) −2−2×((2(2^n −1))/(2−1))  2S_n =S_n +(n−1)2^(n+1) +2  ⇒S_n =(n−1)2^(n+1) +2

Answered by Dwaipayan Shikari last updated on 15/Jul/20

S_n =1.2+2.2^2 +3.2^3 +...........+n.2^n     2S_n =1.2^2 +2.2^3 +.....+(n−1)2^n +n2^(n+1)   ............Subtracting  −S_n =1.2+1.2^2 +1.2^3 +....+2^n −n2^(n+1)   S_n =n2^(n+1) −(1.2+((2^n −1)/1))  S_n =(n−1)^(2n+1) +2  S_(17) =16.2^(18) +2  (⊛■★  DS)

Answered by OlafThorendsen last updated on 15/Jul/20

f(x) = Σ_(k=0) ^n x^k  = ((1−x^(n+1) )/(1−x))  f′(x) = Σ_(k=1) ^n kx^(k−1)   and f′(x) = ((−(n+1)x^n (1−x)+(1−x^(n+1) ))/((1−x)^2 ))  xf′(x) = Σ_(k=1) ^n kx^k  = x((1−(n+1)x^n +nx^(n+1) )/((1−x)^2 ))  ⇒Σ_(k=1) ^(17) k2^k  = 2((1−(17+1)2^(17) +17.2^(17+1) )/((1−2)^2 ))  ⇒Σ_(k=1) ^(17) k2^k  = 2(17.2^(18) −18.2^(17) +1)  ⇒Σ_(k=1) ^(17) k2^k  = 4.194.306