<s>[INST] 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.
 [/INST] 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 $n+1=101,n+1-2r=1$
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