- #1

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## Homework Statement

Prove by induction that the sum of the first n "squares" is n(n+1)(2n+1)/6

In other words prove ([itex]\forall[/itex]n) P(n) where P (n) is [itex]\sum[/itex][itex]^{n}_{i=1}[/itex]¡[itex]^{2}[/itex]=[itex]\frac{n(n+1)(2n+1)}{6}[/itex]

## 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 [itex]\sum[/itex][itex]^{k}_{i=1}[/itex]i[itex]^{2}[/itex]=[itex]\frac{k (k+1)(2k+1)}{6}[/itex] show P (k+1) is true, then ∑[itex]^{k+1}_{i=1}[/itex]i[itex]^{2}[/itex]=[itex]\frac{(k+1)(k+1)(2k+1)}{6}[/itex]

[itex]\sum[/itex][itex]^{k+1}_{i=1}[/itex]i[itex]^{2}[/itex]=[itex]\sum[/itex][itex]^{k}_{i=1}[/itex]i[itex]^{2}[/itex]+(k+1)[itex]^{2}[/itex]

=[itex]\frac{k (k+1)(2k+1)}{6}[/itex] + (k+1)[itex]^{2}[/itex]

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