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Proof by induction that $1^2 + 2^2 + 3^2 +.......+ n^2 = \frac{1}{6}\cdot n\cdot (1+n)\cdot (1+2n)$

I tried showing that

$1^2 + 2^2 + 3^2 +.......+ (k+1)^2 = \frac{1}{6}\cdot (k+1)\cdot (1+(k+1))\cdot (1+2(k+1))$

By using the left side:

$1^2 + 2^2 + 3^2 +.......+ (k+1)^2$

$= 1^2 + 2^2 + 3^2 +.......+ k^2 +(k+1)^2 $

$= \frac{1}{6}\cdot k\cdot (1+k)\cdot (1+2k) + (k+1)^2$

I tried expanding and making it equal the right side but I was not able to get it.

lakada
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2 Answers2

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It is:

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

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

And $2k^2+7k+6=(k+2)(2k+3)$

So we get:

$=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$

And we are done.

Cornman
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The induction step is always the most difficult part, but of course we do need to verify the base case in any induction proof. So I leave it to you to make sure the base case holds (which is trivial but still necessary).

Using the induction hypothesis we have $$\begin{align}1^2+2^2+\cdots+k^2+(k+1)^2&=[1^2+2^2+\cdots+k^2]+(k+1)^2\\&=\frac{k(k+1)(2k+1)}{6}+(k+1)^2\\&=\frac{k(k+1)(2k+1)+6(k+1)^2}{6}\\&=\frac{(k+1)(2k^2+k+6k+6)}{6}\\&=\frac{(k+1)(2k^2+7k+6)}{6}\end{align}$$ I'll leave it to you to factor the final quadratic in the numerator.

Spoiler:

Unsurprisingly we have $2k^2+7k+6=(2k+3)(k+2)$ and thus $$\begin{align}1^2+2^2+\cdots+k^2+(k+1)^2&=\frac{(k+1)(k+2)(2k+3)}{6}\\&=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}\end{align}$$so the formula holds for $k+1$, so we're done.

Dave
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