Question 1000919
{{{1+4+13+40+121}}} to {{{n}}} terms



The first thing you should always do is look at the differences between the terms:

{{{4 - 1 = 3}}}
{{{13 - 4 = 9}}}....and so on.

terms: {{{1}}}, {{{4}}}, {{{13}}}, {{{40}}}, {{{121}}}, ...
differences: {{{3}}}, {{{9}}}, {{{27}}}, {{{81}}}

Now we can see a pattern in the differences: {{{3^n}}} where {{{n}}}={{{1}}},{{{2}}},....and so on, because {{{3^1 = 3}}}, {{{3^2 = 9}}}, {{{3^3 = 27}}}, {{{3^4 = 81}}}

So the next number must be
{{{121 + 3^5 = 121 + 243 = 364}}}

and the next one is
{{{364 + 3^6 = 364 + 729 = 1093}}}

the {{{n}}}th term can be calculated using the formula: 

{{{a[n]=(3^n - 1)/2}}}

and the sum can be calculated using the formula:

{{{1+4+13+40+121}}}+...+{{{(1/2) (3^n-1)}}}

To find the sum of the first n terms of an arithmetic series use the formula,

{{{S[n]=n(a[1]+a[n])/2}}}

{{{S[n]=n(1+(1/2 )(3^n-1))/2}}}

{{{S[n]=n((1+3^n/2 -1/2))/2}}}

{{{S[n]=((n/2+n*3^n/2 ))/2}}}

{{{sum((1/4 )((3^n-1) n+n/2),n=1,n)}}}