how can one find the value of the expression, $(1^2+2^2+3^2+\cdots+n^2)$ [duplicate]

how can one find the value of the expression, $(1^2+2^2+3^2+\cdots+n^2)$

Let,

$T_{2}(n)=1^2+2^2+3^2+\cdots+n^2$

$T_{2}(n)=(1^2+n^2)+(2^2+(n-1)^2)+\cdots$

$T_{2}(n)=((n+1)^2-2(1)(n))+((n+1)^2-2(2)(n-1))+\cdots$

• @RossMillikan but that duplicate doesn't really provide much help for the OP.
– user2468
Aug 30 '12 at 0:14
• I'm trying to locate another question that deals specifically with $1^2 + 2^2 \dots + n^2$. Search sucks!
– user2468
Aug 30 '12 at 0:16
• @JenniferDylan: Robert Israel's answer provides a nice link for powers besides 2 Aug 30 '12 at 0:16
• I downvoted because I thought Rajesh should have caught a clue after his sum-of-cubes question was closed an hour before.
– MJD
Aug 30 '12 at 1:05
• @Jennifer I am sympathetic to the argument that we too often close questions as "duplicates" when they are not, and that closers should take more care to make sure that the similar questions are really close. But in this case I think there are not only one but several exact duplicates.
– MJD
Aug 30 '12 at 1:33

Hint: $(n+1)^3-n^3=3n^2+3n+1$ and use telescopic sum.

In general, $$\sum_{i=1}^{n}i^{2} = \frac{n(n+1)(2n+1)}{6}.$$

A collection of proofs of this fact can be found here.

The claim is that

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

We will verify this by induction.

Clearly $n = 1$ holds.

Suppose the formula holds for $n$. Lets verify it holds for $n + 1$.

$$\sum_{i = 1}^{n + 1} i^2 = \sum_{i = 1}^n i^2 + (n + 1)^2 = \frac{n(n + 1)(2n + 1)}{6} + (n + 1)^2 \\ = \frac{n(n + 1)(2n + 1)}{6} + \frac{6(n^2 + 2n + 1)}{6} \\ = \frac{2n^3 + 3n^2 + n}{6} + \frac{6n^2 + 12 n + 6}{6} \\ = \frac{2n^3 + 9n^2 + 13n + 6}{6}$$

If you factor you get

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

The result follows for $n + 1$. So by induction the formula holds.

• LaTeX tip: MathJax supports  for center-line display math. Also inside  MathJax supports \ for new lines.
– user2468
Aug 30 '12 at 0:40
• @JenniferDylan Thanks. It works a lot more like LaTex than I thought. Aug 30 '12 at 0:50
• $\TeX$ / $\LaTeX$ / MathJax basic tutorial and quick reference
– MJD
Aug 30 '12 at 0:57

If you can find or sketch some 3D blocks, there is a fun geometric proof.

Fix some $n$. If you are doing this with real blocks, $n=3$ or $4$ should be convincing. Let's take $n=4$ for now.

Make a $4\times 4\times 1$ base, laid flat, which of course has volume $4^2$.

Now make a $3\times 3 \times 1$ brick and place it, laid flat, above the base with some corners aligned. Continue in this way up to the top, making smaller squares and always aligning with the same corner. You now have a 3D corner of stairs whose volume is $1^2+\cdots +n^2$.

Now the fun part. Make 5 more of these "stair corners", for a total of 6. These six toys can be turned sideways and upside down, and then pieced together to make an $n\times(n+1)\times(2n+1)$ rectangular solid.