There is a nice generalization. For any positive odd integer $a$ , $\sum_{n = 1}^{a - 1} (-1)^n n^2 = \sum_{n = 1}^{a - 1} \left \lfloor \frac{a + 1}a (-1)^n n^2 \right \rfloor = \frac {a(a - 1)}2.$ To confirm the value of the first summation, notice that $-(2k - 1)^2 + (2k)^2 = 4k - 1,$ so we can reindex the first summation to obtain $\begin{aligned} \sum_{n = 1}^{a - 1} (-1)^n n^2 &= \sum_{k = 1}^{(a - 1)/2} (4k - 1) \\ &= 4 \sum_{n = 1}^{(a - 1)/2} k - \frac {a - 1}2 \\ &= 4 \cdot \frac 12 \cdot \frac {a - 1}2 \cdot \frac {a + 1}2 - \frac {a - 1}2 \\ &= \frac {a(a - 1)}2. \end{aligned}$ Next, for $1 \leq n \leq a - 1$ , write $(-1)^n n^2 = q_n a + r_n$ for integers $q_n$ and $r_n$ with $1 \leq r_n \leq a - 1$ . The fractional part of $\frac{a + 1}a (-1)^n n^2$ is $r_n/a$ , so $\sum_{n = 1}^{a - 1} \left \lfloor \frac {a + 1}a (-1)^n n^2 \right \rfloor = \frac {a + 1} a \sum_{n = 1}^{a - 1} (-1)^n n^2 - \frac 1a \sum_{n = 1}^{a - 1} r_n.$ We have shown that the first summation on the right is equal to $a(a - 1)/2$ . Notice that $\pm k^2 \mp (a - k)^2 \equiv 0 \pmod a$ and thus $r_k + r_{a - k} = a$ . This means $\sum_{n = 1}^{a - 1} r_n = \sum_{k = 1}^{(a - 1)/2} (r_k + r_{a - k}) = \sum_{k = 1}^{(a - 1)/2} a = \frac {a(a - 1)}2.$ Therefore, $\sum_{n = 1}^{a - 1} \left \lfloor \frac {a + 1}a (-1)^n n^2 \right \rfloor = \frac {a + 1}a \cdot \frac {a(a - 1)}2 - \frac 1a \cdot \frac {a(a - 1)}2 = \frac {a(a - 1)}2.$

When $a = 101$ , we obtain $\sum_{n = 1}^{100} \left \lfloor \frac {102}{101} (-1)^n n^2 \right \rfloor = \frac {101 \cdot 100}2 = \boxed{5050}.$

To me, This was a Computer Science question. And i Wrote this code in c++

include<iostream>

using namespace std; int main() { int n,S,i=1,t1=1,t2,term;

cout<<"Enter how much to sum";

cin>>n;

for(i=1;i<=n;i++)
{

t1=t1*(-1);


t2=i*i;


term=102/101 * t2 * t1;

    S=term+S;
}

cout<<"The sum"<<S;

fflush(stdin);

getchar();

return 0;

}