Question 1191490
<br>
The fractional part of the n-th term is<br>
{{{1/(n+1)^2}}}<br>
The other factor in the n-th term is the sum of the cubes of the first n integers:<br>
{{{sum((2n-1)^3,1,n)=((n(n+1))/2)^2}}}<br>
So the nth term in the sequence is<br>
{{{(1/(n+1)^2)(((n(n+1))/2)^2)=(1/(n+1)^2)((n^2/4)(n+1)^2) = n^2/4}}}<br>
There are 51 terms in the sequence, so the sum is<br>
{{{(sum(n^2,1,51)/4)}}}<br>
Use the formula for the sum of the first n squares to get the final answer.<br>
{{{(((51)(52)(103))/6)/4=((51)(52)(103))/24=((17)(13)(103))/2=((221)(103))/2=22763/4}}}<br>
ANSWER: 22763/4<br>