Question 978567


The first term of this geometric progression  {{{a[1]}}} = {{{1}}}, 

and the ratio is  {{{q}}} = {{{-2}}}. 


The formula for the sum of first &nbsp;<B>n</B>&nbsp; terms of a geometric progression is 


{{{S[n]}}} = {{{(a[1]*(q^n-1))/(q-1)}}}


(see everythere, &nbsp;for example, &nbsp;in the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Geometric-progressions.lesson>Geometric progressions</A>&nbsp; in this site). 

Substitute here &nbsp;{{{a[1]=1}}}&nbsp; and &nbsp;&nbsp;<B>n</B>=6, &nbsp;&nbsp;and you will get


{{{S[6]}}} = {{{(1*((-2)^n-1))/(-2-1)}}} = {{{((-1)^n*2^n-1)/(-3)}}} = {{{(1-(-1)^n*2^n)/3}}} = {{{(1 - (-1)^6*2^6)/3}}} = {{{(1 - 2^6)/3}}} = {{{(1-64)/3}}} = {{{-63/3}}} = {{{-21}}}.