$$\sum _{i=2}^{63}\frac{i^{2011}-i}{i^2-1}\bmod 2016$$
How could I do this problem? I know I should be looking for patterns that relate to 2016 and it's multiples, but after that I'm confused.
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Hint:
Using summation formula of Geometric Progression $$\dfrac{n^{2m+3}-n}{n^2-1}=\sum_{r=0}^mn^{2r+1}$$
$$\implies S=\sum_{n=2}^{63}\dfrac{n^{2m+3}-n}{n^2-1}=\sum_{n=2}^{63}\left(\sum_{r=0}^mn^{2r+1}\right)$$
Interchanging the order of summation,
$$S=\sum_{r=0}^m\left(\sum_{n=2}^{63}n^{2r+1}\right)$$
$$=\sum_{r=0}^m\left(\sum_{n=1}^{63}n^{2r+1}\right)-\sum_{r=0}^m1$$
Now $\displaystyle\sum_{r=1}^{63}r=\dfrac{62(63+2)}2\equiv0\pmod{2016}$
Use Show that $1^k+2^k+3^k+ \ldots +n^k$ is divisible by $ 1+2+3+ \ldots +n$
$$\implies S\equiv-\sum_{r=0}^m1\pmod{2016}\equiv2016-(m+1)$$
lab bhattacharjee
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ok I get it, so basically you separate the summation into a piece that you know is divisible by 2016 (so mod 2016 is just zero) and then it's "remainder" component which ends up being the answer – mathkid225 Mar 04 '18 at 05:27
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$\displaystyle\sum_{r=1}^{63}r=\frac {63(64)}2=2016\equiv 0\mod 2016$. – Hypergeometricx Mar 05 '18 at 17:24
Now $\displaystyle\sum_{n=2}^{63}n=\dfrac{62(63+2)}2\equiv-1\pmod{2016}$
– lab bhattacharjee Mar 04 '18 at 05:13