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Question Number 165505 by mathlove last updated on 02/Feb/22

Answered by TheSupreme last updated on 02/Feb/22

Σ_(i=0) ^n (n−i)e^i =n((1−e^(n+1) )/(1−e))−Σie^i = =n((1−e^(n+1) )/(1−e))−(1/e)ΣD(e^i )= =n((1−e^(n+1) )/(1−e))−(1/e)D(((1−e^(n+1) )/(1−e)))= =((n(1−e^(n+1) ))/(1−e))−(1/e)[((−(n+1)(1−e)+(1−e^(n+1) ))/((1−e)^2 ))]= =((ne(1−e^(n+1) )(1−e)−(1−e^(n+1) )+(1−e)(n+1))/(e(1−e)^2 ))= =((ne(1−e^(n+1) −e+e^(n+2) )−1+e^(n+1) +n+1−ne−1)/(e(1−e)^2 ))= =((−ne^(n+2) −ne^2 +ne^(n+3) +e^(n+1) +n−1)/(e(1−e)^2 ))

∑ni=0(n−i)ei=n1−en+11−e−Σiei==n1−en+11−e−1eΣD(ei)==n1−en+11−e−1eD(1−en+11−e)==n(1−en+1)1−e−1e[−(n+1)(1−e)+(1−en+1)(1−e)2]==ne(1−en+1)(1−e)−(1−en+1)+(1−e)(n+1)e(1−e)2==ne(1−en+1−e+en+2)−1+en+1+n+1−ne−1e(1−e)2==−nen+2−ne2+nen+3+en+1+n−1e(1−e)2

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