Question 954448
It is the series of sum of the first {{{n^2}}} terms..

We know that....

Sum of first n term is {{{n(n+1)/4}}}

Sum of the square of the first n term is  {{{(n(n+1)(2n+1))/6}}}


Sum of the cube of the first n numbers is {{{(n^2(n+1)^2)/2}}}

Given that..


{{{1+5+14+30+55+.........}}}

Or

{{{1^2+(1^2+2^2)+(1^2+2^2+3^2)+(1^2+2^2+3^2+4^2)+......}}}{{{(n(n+1)(2n+1))/6}}}.

We can write this series also as--- sum= {{{sum(((n(n+1)(2n+1))/6),n=1,n)}}}.
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Sum={{{sum((((n^2+n)(2n+1))/6),n=1,n)}}}
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Sum={{{sum(((2n^3+3n^3+n)/6),n=1,n)}}}
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Sum={{{sum((((n^3)/3)+((n^2)/2)+(n/6)),n=1,n)}}}
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Sum={{{(1/3)sum(n^3,n=1,n) + (1/2)sum(n^2,n=1,n) + (1/6)sum(n,n=1,n)}}}
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Sum= {{{(1/3)((n^2(n+1)^2)/4) + (1/2)((n(n+1)(2n+1))/6) + (1/6)((n(n+1))/2)}}}

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Sum= {{{(n(n+1)(n(n+1)+(2n+1)+1))/12}}}

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Sum= {{{(n(n+1)(n^2+3n+2))/12}}}

.

Sum= {{{(n(n+1)(n+1)(n+2))/12}}}

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Sum= {{{(n(n+1)^2(n+2))/12}}}.... Ans..

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To varify this formula lets take n=4... 
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Sum={{{(4(4+1)^2(4+2))/12}}}
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Sum={{{(4*25*6)/12}}}
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Sum= 50.
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In the series...  Sum of first 4 terms= 1+5+14+30 = 50..
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Which is equal to the our calculated Ans....