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Question-208322




Question Number 208322 by alcohol last updated on 11/Jun/24
Answered by Berbere last updated on 11/Jun/24
Π_(k=1) ^(n−1) (1+(1/k))^k =Π_(k=1) ^(n−1) (((1+k)^k )/k^k )=Π_(k=1) ^(n−1) (((1+k)^(k+1) )/((k+1).k^k ))  =Π_(k=1) ^(n−1) (1/((k+1))).Π_(k=1) ^(n−1) (((1+k)^(k+1) )/k^k )=(1/(n!)).n^n =(n^(n−1) /((n−1)!))
n1k=1(1+1k)k=n1k=1(1+k)kkk=n1k=1(1+k)k+1(k+1).kk=n1k=11(k+1).n1k=1(1+k)k+1kk=1n!.nn=nn1(n1)!
Answered by MathematicalUser2357 last updated on 13/Jun/24
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