Question 979194


The first term of this geometric progression  {{{a[1]}}} = 1,   and the ratio is  q = 4. 


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;{{{q=4}}}&nbsp; and &nbsp;&nbsp;<B>n</B>=6, &nbsp;&nbsp;and you will get


{{{S[6]}}} = {{{(4^6-1)/(4-1)}}} = {{{(4^3-1)*(4^3+1)/3}}} = {{{(64-1)*(64+1)/3}}} = {{{(63*65)/3}}} = 21*65 = 1365 .