# Proof by induction

## Homework Statement

Prove by induction that the sum of the first n "squares" is n(n+1)(2n+1)/6
In other words prove ($\forall$n) P(n) where P (n) is $\sum$$^{n}_{i=1}$¡$^{2}$=$\frac{n(n+1)(2n+1)}{6}$

## Homework Equations

This is just not clicking for me right now. I have no idea if i am just epic failing at factoring or doing something else wrong.

## The Attempt at a Solution

I did the base case and proved P (1) true easily enough.
Induction case: suppose P (k), then $\sum$$^{k}_{i=1}$i$^{2}$=$\frac{k (k+1)(2k+1)}{6}$ show P (k+1) is true, then ∑$^{k+1}_{i=1}$i$^{2}$=$\frac{(k+1)(k+1)(2k+1)}{6}$

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

Now i know i need to get itto work out to the $\frac{(k+1)(k+1)(2k+3)}{6}$ but i cant manage to get itto work out at all.

LCKurtz
Homework Helper
Gold Member

## Homework Statement

Prove by induction that the sum of the first n "squares" is n(n+1)(2n+1)/6
In other words prove ($\forall$n) P(n) where P (n) is $\sum$$^{n}_{i=1}$¡$^{2}$=$\frac{n(n+1)(2n+1)}{6}$

## Homework Equations

This is just not clicking for me right now. I have no idea if i am just epic failing at factoring or doing something else wrong.

## The Attempt at a Solution

I did the base case and proved P (1) true easily enough.
Induction case: suppose P (k), then $\sum$$^{k}_{i=1}$i$^{2}$=$\frac{k (k+1)(2k+1)}{6}$ show P (k+1) is true, then ∑$^{k+1}_{i=1}$i$^{2}$=$\frac{(k+1)(k+1)(2k+1)}{6}$

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

Now i know i need to get it to work out to the $\frac{(k+1)(k+1)(2k+3)}{6}$ but i cant manage to get itto work out at all.

You mean you need to work it out to ##\frac{(k+1)(k+\color{red}2)(2k+3)}{6}##. You just need to continue with the algebra. Add up those two terms and simplify it by factoring out ##k+1##.

1 person
Oh gees ya thats probably what my problem was, thank you sometimes you just need another set of eyes.