help 24

This should help you 

https://www.youtube.com/watch?v=L_jWHffIx5E

I'm providing this long answer because I believe we should help each other; If you do too, and want to create a better world, VOTE BERNIE SANDERS!

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+---+2^n-1)

= n+ (n-1)2¹+(n-2)2²+(n-3)2³+---+ (n-k)2^k

^{\(=\sum_{k=0}^{n} (n-k) 2^{k} \\ = n \sum_{k=0}^{n} 2^{k} - \sum_{k=0}^{n} k \ 2^{k} \\ \\ \mbox{ (now, since} \sum_{k=m}^{n} r^{k} = \frac{r^{m} - r^{n+1}}{1-r} \\ \frac{d}{dr}\sum_{k=0}^nr^k = \sum_{k=1}^n kr^{k-1}= \frac{1-r^{n+1}}{(1-r)^2}-\frac{(n+1)r^n}{1-r} )\\ \mbox{we get:} \\ = n (2^{n+1}-1) - 2[(1-2^{n+1}) + 2^{n}(n+1)] \\ \\ \mbox{and, of course, you can simplify, and test for cases.} \)}

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.

(1+3+7+15+31+..........+n-th term), it is a G.P series. Now find the sumtotal upto n-th term.

This is NOT a GP series.

Guest #2 seems to know whats what.   (I'm not talking about U.S. politics either)

This is NOT a G.P., but you can sum it up as follows:

2F*{[1 - r^n] / [1 - r ]} - n        F=1st.term, r=ratio, n=number of terms.

2*1{[1 - 2^10] / [1 - 2 ]} - 10

2{[1 - 1024] / - 1 } - 10

2*{-1023/-1 } - 10

2*1023 - 10

2046 -10

=2,036 this is the sum of the 1st. 10 terms.