Prove by induction that $(n+1)^2 + (n+2)^2 + ... + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}$

Question

Prove by induction that $$(n+1)^2 + (n+2)^2 + ... + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}.$$ I got up to: $n=1$ is true, and assuming $n=k$ prove for $n=k+1$.

Prove...

$$\frac{k(2k+1)(7k+1)+6(2(k+1))^2}{6} = \frac{(k+1)(2k+3)(7k+8)}{6}$$

I keep trying to expand to $6(2k+2)^2$ and factorising but I end up being short on one factor, e.g., I end up with $\frac{(k+1)(2k+3)(7k+6)}{6}$.

Answer 1

Hint :

Be careful, at the step $k+1$ the sum is $$ (k+2)^2+\cdots+(2k+1)^2+(2k+2)^2 $$ and the sum at the step $k$ is $$ (k+1)^2+\cdots+(2k-1)^2+(2k)^2 $$

Answer 2

Alternative route (also with induction)

You can start proving inductively that: $$\sum_{k=1}^{n}k^{2}=\frac{1}{6}n\left(n+1\right)\left(2n+1\right)$$ Then have a look at: $$\sum_{k=n+1}^{2n}k^{2}=\sum_{k=1}^{2n}k^{2}-\sum_{k=1}^{n}k^{2}$$

Answer 3

Check this out # $(k+1)^2 +(k+2)^2+ ............ (k+k)^2 = \frac{k(2k+1)(7k+1)}{6}$

# $p(k+1): (k+1)^2+(k+2)^2...........(2k)^2+(2k+1)^2+.(2k+2)^2-(k+1)^2= \frac{(k+1)(2k+3)(7k+8)}{6}$

# $p(k+1): \frac{k(2k+1)(7k+1)}{6}+3(k+1)^2+(2k+1)^2$=RHS

Answer 4

Inductive step spelled out in detail:

\begin{align*} &\hphantom{=}\frac{n(2n+1)(7n+1)}6+(2n+1)^2+(2n+2)^2-(n+1)^2\\ &=\frac{n(2n+1)(7n+1)}6+(2n+1)^2+3(n+1)^2\\ &=(2n+1)\left(\frac{n(7n+1)}6+2n+1\right)+3(n+1)^2\\ &=(2n+1)\frac{n(7n+1)+12n+6}6+3(n+1)^2\\ &=(2n+1)\frac{7n^2+13n+6 }6+3(n+1)^2\\ &=\frac{(2n+1)(n+1)(7n+6)}6+3(n+1)^2\\ &=(n+1)\left(\frac{(2n+1)(7n+6)}6+3(n+1)\right)\\ &=(n+1)\frac{(2n+1)(7n+6)+18(n+1)}6\\ &=(n+1)\frac{14n^2+19n+6+18n+18}6\\ &=(n+1)\frac{14n^2+37n+24}6\\ &=\frac{(n+1)(2n+3)(7n+8)}6 \end{align*}