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I have to prove that this statement is true.

For $n = 1, 2, 3, ...,$ we have $ 1² + 2² + 3² + ... + n² = n(n + 1)(2n + 1)/6$

Basically I thought I'd use induction to prove this. Setting $n = p+1$, I got this so far:

Left hand side: $1² + 2² + 3² +...+ p² + (p+1)² = p² + (p+1)² = 2p² + 2p + 1$

Right hand side: $(p + 1)(p + 1 + 1)(2 * (p + 1) +1)/6$

This is all I've got. I've never worked with induction so I don't even know if this correct to begin. Any insight or solution to my problem is appreciated.

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    $\begingroup$ $1^2 + 2^2 + \cdots + p^2 + (p+1)^2 = p^2 + (p+1)^2$? $\endgroup$ – 6005 May 19 '13 at 18:14
  • $\begingroup$ I guess thats what I'm looking for, am I using induction right though? $\endgroup$ – Simon Carlson May 19 '13 at 18:22
  • $\begingroup$ It is rather hard to tell because your question is confusing and contains some false statements that you probably don't mean. I'll post an answer to explain. $\endgroup$ – 6005 May 19 '13 at 18:24
  • $\begingroup$ Thank you. I don't know how to express myself when talking about math, even less in English. $\endgroup$ – Simon Carlson May 19 '13 at 18:25
  • $\begingroup$ @SimonCarlson: Look at my answer. Do you understand it? $\endgroup$ – Inceptio May 19 '13 at 18:29
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For proving an expression through induction. Check for the base case $n=1$,

Let $P(n)$ is true, $\sum_1^nk^2=1+4+9+ \dots n^2= P(n)=\dfrac{n(n+1)(2n+1)}{6}$.

You need to prove that $\sum_1^{n+1} (k)^2=1+4+9+ \dots (n+1)^2=P(n+1)=\dfrac{{(n+1)}{}(n+2)(2n+3)}{6}.$

Hint: $(k+1)^2=k^2+2k+1$ and $\sum_1^n k=\dfrac{n(n+1)}{2}, \sum_1^n 1=n$

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  • $\begingroup$ One quibble: you have $P(k)$ here denoting a statement (the induction hypothesis) as well as an expression ($1^2+\cdots+k^2$). $\endgroup$ – Shane O Rourke May 19 '13 at 19:16
  • $\begingroup$ @Shane, the statement hasn't been proved. Do you know notice You need to prove that there? Btw, thanks for the parentheses. $\endgroup$ – Inceptio May 19 '13 at 19:18
  • $\begingroup$ A further quibble: the roles of $k$ and $n$ appear to be mixed up: there's an $n$ on the left-hand side of your first equation but only a $k$ on the right-hand side. Also, the left-hand side of the `$P(k+1)$' statement should be something like $\sum_{m=1}^{k+1}m^2$, not $\sum (k+1)^2$ $\endgroup$ – Shane O Rourke May 19 '13 at 19:21
  • $\begingroup$ Yes ,ShaneORourke. I got them edited. More quibbles? $\endgroup$ – Inceptio May 19 '13 at 19:22
  • $\begingroup$ Agreed that $P(k)$ has not been proven (at the point it's stated) and that $P(k+1)$ is a statement of what we need to prove. And agreed that it's often a good idea to spell these things out. My (first!) quibble was only that the notation $P(k)$ is playing two roles. $\endgroup$ – Shane O Rourke May 19 '13 at 19:25
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First of all, your statement $1^2+2^2+⋯+p^2+(p+1)^2=p^2+(p+1)^2$ is false. You have simply erased $1^2 + 2^2 + \cdots + (p-1)^2$ from the equation, which you really can't do.

It seems like you are trying to plug in $(p+1)$ to the LHS and the RHS of the equation. That's fine, but now what you have to do is show that if the LHS and RHS side are equal when $n = p$, they are also equal when $n = p+1$. So, you've calculated the new LHS and RHS for $n = p+1$ -- now show they're equal.

Inceptio's answer should be a sufficient hint at this point.

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I think you mean $$\begin{array}{rcl}1^2+2^2+\cdots+p^2+(p+1)^2 &=& \sum_{k=1}^p k^2+(p+1)^2\\ & = & \frac{p(p+1)(2p+1)}{6}+(p+1)^2\end{array}$$

In other words by writing just $1^2+\cdots +p^2+(p+1)^2=p^2+(p+1)^2$, the first $p-1$ terms from the left-hand side are missing, so it's not a valid equality.

At any rate, tidying up the displayed equation above, you should get what you need.

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