Find value of $a_1$ such that $a_{101}=5075$ Let $\{a_n\}$ be a sequence of real numbers where $$a_{n+1}=n^2-a_n,\, n=\{1,2,3,...\}$$
Find value of $a_1$ such that $a_{101}=5075$.

I have 
$$a_2=1^2-a_1$$
$$a_3=2^2-a_2=2^2-1^2+a_1$$
$$a_4=3^2-2^2+1^2-a_1$$
$$a_5=4^2-3^2+2^2-1^2+a_1$$
$$\vdots$$
$$a_{101}=100^2-99^2+98^2-97^2+\ldots+2^2-1^2+a_1.$$
Therefore, $$a_{101}=\sum_{i=1}^{50}(2i)^2-\sum_{i=1}^{50}(2i-1)^2.$$
Thus, $$5075=\sum_{i=1}^{50}(4i^2-4i^2+4i-1)+a_1,$$
and, $$a_1=5075-4\sum_{i=1}^{50}(i)+\sum_{i=1}^{50}(1).$$
Hence, $$a_1=5075-4(\frac{50}{2})(51)+50=25,$$
and $a_1=25$.
Is it correct? Do you have another way? Please check my solution, thank you.

What if at the point where:
$a_{101} = 100^2 - 99^2 + 98^2 - 97^2 \cdots 2^2 - 1^2 + a_1$
You opted for a clever factorization of squares:
$a_{101} = (100 - 99)(100 + 99) + (98 - 97)(98 + 97) \cdots (2 - 1)(2 + 1) + a_1$
$a_{101} = 199 + 195 + 191 \cdots 3 + a_1$
Therefore those numbers become a simple arithmetic series with the first term as 3 and the common ratio as 4.
But how many terms exactly?
$3 + 4(n - 1) = 199$
$4(n - 1) = 196$
$n - 1 = 49$
$n = 50$
Therefore with the knowledge of evaluating the sum this comes down to:
$a_{101} = 25(199 + 3) + a_1$
$a_{101} = 5050 + a_1$
And by substituting the choice for a_101:
$5075 = 5050 + a_1$
$a_1 = 25$
Thanks to @Zera for recommending the edit. I'm learning how to use these markup languages
