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.

Yes, your solution is correct.  Another method of solution is to note that if $$a_{n+1} = n^2 - a_n,$$ we want to find some (possibly constant) function of $n$ such that $$a_{n+1} - f(n+1) = -(a_n - f(n)).$$  This of course implies $$f(n+1) + f(n) = n^2.$$  A quadratic polynomial should do the trick:  suppose $$f(n) = an^2 + bn + c,$$ so that $$n^2 = f(n+1) + f(n) = 2a n^2 + 2(a+b)n + (a+b+2c).$$  Equating coefficients in $n$ gives $a = 1/2$, $b = -1/2$, $c = 0$, hence $$f(n) = \frac{n^2 - n}{2}.$$  It follows that if $$b_n = a_n - f(n) = a_n - \frac{n^2-n}{2},$$ then $$b_{n+1} = - b_n.$$  This gives us $$b_1 = b_{101}$$ which in terms of $a_n$, is $$a_1 = a_1 - \frac{1^2 - 1}{2} = a_{101} - \frac{(101)^2 - 101}{2} = 5075 - 5050 = 25.$$
This solution seems to come out of nowhere, but it is motivated by the idea that if we can transform the given recurrence into a corresponding recurrence for a sequence that is much simpler to determine, we can use this to recover information about the original sequence.
