# Proof by induction

1. Mar 3, 2014

### mikky05v

1. The problem statement, all variables and given/known data

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}$

2. Relevant 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.

3. 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.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 3, 2014

### LCKurtz

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 $k+1$.

3. Mar 4, 2014

### mikky05v

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

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