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.
 A: 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.
A: 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
A: In the most simple way:
$$A_{n+1}+A_n=n^2~~~(1)$$ is a non-homogeneous recurrence equation. Its homogeneous part
$$A_{n+1}+A_n=0 \implies A_{n+1}=-A_n \implies A_n= (-1)^n S~~~(2)$$ In the RHS of (1) being $n^2)$, we can take $A_n =P n^2+Q n+R$; inserting this in (1), we get
$2P=1,P+Q=0,P+Q+2R=0 \implies P=1/2, Q=1/2, R=0.$ Then the total finally solution of (1) is
$$A_n=\frac{n(n-1)}{2}+ (-1)^n S. $$  Given that $A_{101}=5075$, finally we get
$$A_n=\frac{n(n-1)}{2}+(-1)^{n+1}~ 25.$$
A: Another way :
$$a_{n+1}=n^2-a_n=n^2-((n-1)^2-a_{n-1})=a_{n-1}+2n-1$$
$$a_{n-1}=a_{n-3}+2(n-1)-1=a_{n-3}+2(n-2)+1$$
$$a_{n+1}=a_{n+1-2r}+\underbrace{2n-1+2n-3+\cdots}_{r\text{ terms}}$$
Here $n+1=101,n+1-2r=1$
A: Your solution is nice.
You could also have notice the pattern
$$a_n=(-1)^{n+1}a_1+\frac{n(n-1) }{2} $$
