Solution

Given equation is

$\left(\sum_{k=1}^{n}(-1)^{k-1}k\right)^2-\sum_{k=1}^{n}(-1)^{k-1}k^2+2450=0$

Check for odd $n$.

Let

$n=2m-1$

Then,

$\sum_{k=1}^{n}(-1)^{k-1}k=1-2+3-4+\cdots+(2m-1)$

Pairing terms,

$(1-2)+(3-4)+\cdots+[(2m-3)-(2m-2)]+(2m-1)$

$=-(m-1)+(2m-1)$

$=m$

So,

$\left(\sum_{k=1}^{n}(-1)^{k-1}k\right)^2=m^2$

Now,

$\sum_{k=1}^{n}(-1)^{k-1}k^2=1^2-2^2+3^2-4^2+\cdots+(2m-1)^2$

For odd $n=2m-1$,

$\sum_{k=1}^{n}(-1)^{k-1}k^2=m(2m-1)$

Now put in the equation:

$m^2-m(2m-1)+2450=0$

$m^2-2m^2+m+2450=0$

$-m^2+m+2450=0$

$m^2-m-2450=0$

Now factorize:

$m^2-m-2450=0$

$m^2-50m+49m-2450=0$

$m(m-50)+49(m-50)=0$

$(m-50)(m+49)=0$

Since $m$ is positive,

$m=50$

Therefore,

$n=2m-1$

$n=2(50)-1$

$n=99$