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Question Number 147697 by mathdanisur last updated on 22/Jul/21
f(0)+f(1)+f(2)+...+f(n)=n!−n⋅af(16)=15⋅(15!−1)finda=?
Answered by Olaf_Thorendsen last updated on 22/Jul/21
∑nk=0f(k)=n!−na(1)⇒∑n+1k=0f(k)=(n+1)!−(n+1)a(2)(2)−(1):f(n+1)=(n+1)!−(n+1)a−n!+naf(n+1)=n.n!−aIfn=15:f(16)=15.15!−af(16)=15(15!−a15)=15(15!−1)⇒a=15
Commented by mathdanisur last updated on 22/Jul/21
ThankyouSer
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