Given that $\sum\limits_{r=1}^{n}{{{r}^{2}}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$. Find

Question

Given that $\sum\limits_{r=1}^{n}{{{r}^{2}}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$. Find, in terms of $n$, $\sum\limits_{r=-2}^{n}{{{r}^{2}}}$ $\sum\limits_{r=-3}^{n+1}{{{r}^{2}}}$ $\sum\limits_{r=-2}^{n-3}{2{{r}^{2}}}$

Accepted Answer

(i) $5+\frac{n(n+1)(2n+1)}{6}$ (ii) $14+\frac{(n+1)(n+2)(2n+3)}{6}$ (iii) $10+\frac{(n-3)(n-2)(2n-5)}{3}$

Change of Base Q2

Change of Base Q2

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