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Question Number 151097 by naka3546 last updated on 18/Aug/21
Findsumofthisexpression.(n1)+2(n2)+3(n3)+4(n4)+…+n(nn)Pleaseshowyourworkings.Thankyou.
Answered by Olaf_Thorendsen last updated on 18/Aug/21
Sn=∑nk=1kCknf(x)=(x+1)n=∑nk=0Cknxkf′(x)=n(x+1)n−1=∑nk=1kCknxk−1f′(1)=n2n−1=∑nk=1kCkn=Sn
Answered by mr W last updated on 18/Aug/21
Sn=(n1)+2(n2)+3(n3)+4(n4)+…+n(nn)Sn=0(n0)+(n1)+2(n2)+3(n3)+4(n4)+…+n(nn)Sn=n(nn)+(n−1)(nn−1)+(n−2)(nn−2)+(n−3)(nn−3)+…+0(nn−n)Sn=n(n0)+(n−1)(n1)+(n−2)(n2)+(n−3)(n3)+…+0(nn)2Sn=n[(n0)+(n1)+(n2)+(n3)+…+(nn)]2Sn=n×2n⇒Sn=n×2n−1
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