Question 944648

sequence {{{4}}},{{{8}}},{{{16}}},{{{32}}},{{{64}}},{{{128}}}…

{{{d=2}}} ---> common multiplier 

{{{a}}} sub {{{1}}}={{{4}}} => {{{a}}} sub {{{1}}}={{{a[1]}}}  which is first term in sequence and first term is {{{4}}}

prob #1:{{{ a[3]}}}  => third term which is {{{16}}}; so, {{{a[3] =16}}}

prob #2:{{{ a[1]+a[7] =4+a[7] }}}

 => we need general formula to find {{{a[7]}}}

sequence {{{4}}},{{{8}}},{{{16}}},{{{32}}},{{{64}}},{{{128}}}…using common multiplier could be written as {{{2^2}}},  {{{2^3}}},{{{ 2^4}}}, {{{2^5}}}, {{{2^6}}}..... {{{2^(n+1)}}}

so, general formula is:

{{{a[n] = 2^(n+1)}}} (for all terms given)  where {{{n}}} is the number of the term

now find {{{a[7]}}} 

{{{a[7] = 2^(7+1) }}}

{{{a[7] = 2^8}}}

{{{a[7] = 256}}}

and {{{ a[1]+a[7] =4+a[7] }}} will be

{{{ a[1]+a[7] =4+256 }}}

{{{ a[1]+a[7] =260 }}}