Question 1161218


it's {{{a[n]=2^n+3}}} where {{{n}}}={{{1}}},{{{2}}},{{{3}}}........

check:

{{{n=1}}}
{{{a[1]=2^1+3=2+3=5}}}

{{{n=2}}}
{{{a[2]=2^2+3=4+3=7}}}

{{{n=3}}}
{{{a[3]=2^3+3=8+3=11}}}


{{{n=4}}}
{{{a[4]=2^4+3=16+3=19}}}


{{{n=5}}}
{{{a[5]=2^5+3=32+3=35}}}