Question 927581
{{{5 + 20 + 80 }}}+ ....+ {{{20480}}}

{{{a[1]=5}}}...first term
if you multiply it by {{{4}}} you got second term {{{20}}}, then if you multiply it by {{{4}}} you got third term {{{80}}}

if we write {{{4}}} as {{{2^2}}}, but to get first term we can write {{{2^(2n-2)}}}

so, general formula is {{{a[n]=5*2^(2n-2)}}}

now find which term is {{{20480}}}

{{{20480=5*2^(2n-2)}}}

{{{20480/5=2^(2n-2)}}}

{{{4096=2^(2n-2)}}}

{{{2*2*2*2*2*2*2*2*2*2*2*2=2^(2n-2)}}}

{{{2^12=2^(2n-2)}}}...if bases same, then

{{{12=2n-2}}} ...solve for {{{n}}}

{{{12+2=2n}}}

{{{14=2n}}}

{{{7=n}}}

also find:

{{{a[4]=5*2^(2*4-2)}}}

{{{a[4]=5*2^(6)=320}}}


{{{a[5]=5*2^(2*5-2)=5*2^(8)=1280}}}

{{{a[6]=5*2^(2*6-2)=5*2^(10)=5120}}}


then the sum is:

{{{sum(5*2^(2n-2),n=1,7)=5+20+80+320+1280+5120+20480 =27305}}}