Question Number 100236 by mathmax by abdo last updated on 25/Jun/20

calculate Σ_(k=0) ^n  (C_n ^k /((k+1)^2 ))

Answered by mathmax by abdo last updated on 27/Jun/20

let f(x) =Σ_(k=0) ^n  (C_n ^k /(k+1)) x^k  ⇒ ∫_0 ^x  f(t)dt =Σ_(k=0) ^n  (C_n ^k /((k+1)^2 )) x^(k+1)   let explicite f(x)  we have  Σ_(k=0) ^n  C_n ^k  x^(k ) =(x+1)^(n )  ⇒  Σ_(k=0) ^n  (C_n ^k /(k+1))x^(k+1)  =(1/(n+1))(x+1)^(n+1)  +c  x=0 ⇒0 =(1/(n+1))+c ⇒c =−(1/(n+1)) ⇒Σ_(k=0) ^n  (C_n ^k /(k+1))x^(k+1)  =(((x+1)^(n+1) −1)/(n+1)) ⇒  f(x) =(((x+1)^(n+1) −1)/((n+1)x)) ⇒Σ_(k=0) ^n  (C_n ^k /((k+1)^2 ))x^(k+1)  =(1/(n+1))∫_0 ^x  (((t+1)^(n+1) −1)/t)dt  ⇒Σ_(k=0) ^n  (C_n ^k /((k+1)^2 )) =(1/(n+1)) ∫_0 ^1  (((t+1)^(n+1) −1)/t)dt  we have  (t+1)^(n+1) −1 =t(1+(t+1)+(t+1)^2 +.....(t+1)^n ) ⇒  (((t+1)^(n+1) −1)/t) =1 +(t+1)+(t+1)^2  +....(t+1)^n   ⇒  ∫_0 ^1  (((t+1)^(n+1) −1)/t)dt =∫_0 ^1 (1+(t+1)+(t+1)^2  +...+(t+1)^n )dt  =[t +(((t+1)^2 )/2) +(((t+1)^3 )/3) +.....(((t+1)^(n+1) )/(n+1))]_0 ^1   =1+(2^2 /2) +(2^3 /3) +.....+(2^(n+1) /(n+1))−(1/2)−(1/3)−....−(1/(n+1)) ⇒  Σ_(k=0) ^n  (C_n ^k /((k+1)^2 )) =(1/(n+1)){ 2+(2^2 /2) +(2^3 /3)+....+(2^(n+1) /(n+1))−H_(n+1) }