Is it AP?

This can be looked at as two sequences with $15$ terms each, one involving the terms preceded by the $+$ signs, (including $45^{2}$ ), and one preceded by the $-$ signs. This gives us

$\displaystyle\sum_{n=31}^{45} n^{2} - \sum_{n=29}^{43} n^{2}.$

The overlapping terms of these two sums then cancel, leaving us with just the first two terms of the first sum and the last two of the second. The desired answer is then

$45^{2} + 44^{2} - 29^{2} - 30^{2} = \boxed{2220}.$

Using identity $a^2-b^2=(a+b)(a-b)$ . $(45+43)(45-43)+(44+42)(44-42)+(43+41)(43-41)+......$ $88×2+86×2+84×2+.......$

Now, taking $2$ as common.

$2×(88+86+84+82+.....)$

Now using sum formula.

$\frac{n}{2}×(2a+(n-1)d)$

$=\dfrac{15}{2}×(2×88+14×(-2)$ .

$=\dfrac{15}{2}×148$

$=1110$

Now, sum becomes

$=2×1110$

$\boxed{2220}$