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$$1^2+2^2+3^2+...+10000$$

How do you find the exact value of that?

I'm studying induction, and I'm still not sure how to get that value.

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A related question. –  J. M. Jul 5 '12 at 5:53

3 Answers 3

Let $$S=1+2+3+...+n$$ then, $$S=n+(n-1)+(n-2)+...+1$$ Adding corresponding terms of two equations gives , $$2S=(n+1)+(n+1)+(n+1)+...(n+1)=n(n+1)\implies S=n(n+1)/2$$ Now $$(n+1)^3-n^3=3n^2+3n+1$$ $$\implies 2^3-1^3=3(1)^2+3(1)+1$$ $$3^3-2^3=3(2)^2+3(2)+1$$ $$4^3-3^3=3(3)^2+3(3)+1$$ $$..$$$$..$$$$(n+1)^3-n^3=3(n)^2+3(n)+1$$ Adding corresponding terms gives, $$(n+1)^3-1^3=3(1^2+2^2+...+n^2)+3(1+2+...+n)+n$$ $$\implies n^3+3n^2+3n=3(ReqSum)+3n(n+1)/2+n$$ Solving for $ReqSum$ and factorization gives $ReqSum=n(n+1)(2n+1)/6$. For $n=100$, it is $100.101.201/6=338350$.

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The sum of squares formula is

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

Clearly it holds for $n = 1$.

Suppose it held for $n$. That is

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

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

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

Thus by induction the formula holds.

Your series above is $\sum_{i = 1}^{100} i^2 = \frac{(100)(101)(201)}{6}$

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Some useful formulae, which are easy to prove by induction are:

$$\sum_{i= 1}^n 1 = n$$

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

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

$$\sum_{i= 1}^n \frac{i(i+1)(i+2)}{6} = \frac{n(n + 1)(n+2)(n+3)}{24}$$

In fact, there is a general form for this kind of expression which can be established by induction. This enables us to see that the sum of a quadratic expression will be a cubic one, the sum of a cubic expression will be quartic etc. It also gives a brute force way of summing any such formula - but there are often short cuts. I have found it useful to know the formulae for the sums of squares and cubes.

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+1: Might also want to point out that your formulas follow easily from Pascal's triangle :-). –  Jyrki Lahtonen Jul 5 '12 at 6:22

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