(1+3+7+15+31+..........+n-th term), it is a G.P series. Now find the sumtotal upto n-th term.
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I left out a few steps, but you can fill them in.
1+3+7+15+---+nth => 1 + (1 + 21) + (1+21+22) + (1+21+ 22+23) +(1+21+ 22+23+---+2n-1)
= n+ (n-1)21+(n-2)22+(n-3)23+---+ (n-k)2k
=∑nk=0(n−k)2k=n∑nk=02k−∑nk=0k 2k (now, since∑nk=mrk=rm−rn+11−rddr∑nk=0rk=∑nk=1krk−1=1−rn+1(1−r)2−(n+1)rn1−r)we get:=n(2n+1−1)−2[(1−2n+1)+2n(n+1)]and, of course, you can simplify, and test for cases.
Yo can't sum up a GP to the "nth term". You have to specify what the "nth term" is. Is it the 10nth term, is it the 20th term, the 30th term........ and so on.