Question

answer it fast guys​

Answer 1

sigma Sk=1/(k+1)^i...where i is from 0 to infinity

Sk=1/(k+1)^0 +1/(k+1)^1+1/(k+1)^2.........n(trending to infinity)

Sk=1{1-(1/k+1)^(n+1)/{1-1/(k+1)}

solving this

we get

sigmaSk=(k+1)(1-1/k+1)^(n+1)×k

if k >0 then,(1-(1/k+1)^(n+1) will be negligible

so

Sk=(k+1)/k

sigmaSk( from k=0 to n)=(k+1)/k

kSk={(k+1)/k}×k

KSk(k from 0 to n(=(k+1)

using the method of series

we get

KSk(k from 0 to n)=n(n+1)/2 + n

=n(3+n)/2

=(3n+n^2)/2

hope it helps