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Question Number 100236 by mathmax by abdo last updated on 25/Jun/20
calculate Σ_(k=0) ^n (C_n ^k /((k+1)^2 ))
calculate∑k=0nCnk(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) }
letf(x)=∑k=0nCnkk+1xk⇒∫0xf(t)dt=∑k=0nCnk(k+1)2xk+1letexplicitef(x)wehave∑k=0nCnkxk=(x+1)n⇒∑k=0nCnkk+1xk+1=1n+1(x+1)n+1+cx=0⇒0=1n+1+c⇒c=−1n+1⇒∑k=0nCnkk+1xk+1=(x+1)n+1−1n+1⇒f(x)=(x+1)n+1−1(n+1)x⇒∑k=0nCnk(k+1)2xk+1=1n+1∫0x(t+1)n+1−1tdt⇒∑k=0nCnk(k+1)2=1n+1∫01(t+1)n+1−1tdtwehave(t+1)n+1−1=t(1+(t+1)+(t+1)2+…..(t+1)n)⇒(t+1)n+1−1t=1+(t+1)+(t+1)2+….(t+1)n⇒∫01(t+1)n+1−1tdt=∫01(1+(t+1)+(t+1)2+…+(t+1)n)dt=[t+(t+1)22+(t+1)33+…..(t+1)n+1n+1]01=1+222+233+…..+2n+1n+1−12−13−….−1n+1⇒∑k=0nCnk(k+1)2=1n+1{2+222+233+….+2n+1n+1−Hn+1}

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