Show, with induction that $1^2 + 2^2 + .... + n^2 = \frac{n(n+1)(2n+1)}{6}$

Question

Show, with induction that

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

My attempt

Case 1: n = 1

$LHS = 1^2$

$RHS = \frac{(1+1)(2+1)}{6} = \frac{2*3}{6} = 1$

Case 2: n = p

$LHS_{p} = 1^2 + 2^2 + ... + p^2$

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

Case 3: n = p + 1

$LHS_{p+1} = 1^2+2^2+....+p^2+(p+1)^2$

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

Now to show this with induction I think i need to show that

$RHS_{p+1} = RHS_{p} + (p+1)^2$

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

So I need to rewrite

$RHS_{p+1} = \frac{(p+1)((p+1)+1)(2(p+1)+1)}{6} $to be equal to $\frac{p(p+1)(2p+1)}{6} + (p+1)^2$ Anyone see how I can do that? Or got any other solution?

Simplify both expressions (expand, collect, etc.), or equivalently simplify the difference, which should equal $0$. You could also compute the sum using the hockey-stick identity. — Jean-Claude Arbaut, Apr 15 '19 at 19:05
You may find it helpful to read an answer I wrote a good while back concerning this. It would be helpful to write your proof a little more neatly. — Daniel W. Farlow, Apr 15 '19 at 19:09
There is also https://math.stackexchange.com/questions/48080/sum-of-first-n-squares-equals-fracnn12n16 — Arnaud D., Apr 15 '19 at 19:09

score 0 · Answer 1 · answered Apr 15 '19 at 19:09

0

HINT

In your notation, $$ \begin{split} 1^2+\ldots+p^2+(p+1)^2 &= \frac{p(p+1)(2p+1)}{6} + (p+1)^2 \\ &= \frac{p+1}{6} \left[p(2p+1)+6(p+1)\right] \\ &= \frac{p+1}{6} \left[2p^2 + 8p+6\right] \\ &= \frac{(p+1)(p+2)(2p+3)}{6} \end{split} $$

answered Apr 15 '19 at 19:09

gt6989b

54,422

Hey thanks man :) – Daniel Andersson Apr 15 '19 at 19:17

Show, with induction that $1^2 + 2^2 + .... + n^2 = \frac{n(n+1)(2n+1)}{6}$

1 Answers1