User:Temperal/The Problem Solver's Resource3: Difference between revisions

Revision as of 21:52, 10 January 2009

Introduction	Other Tips and Tricks	Methods of Proof	You are currently viewing page 3.

Summations: $\sum_{i=a}^{b}c_i=c_a+c_{a+1}+c_{a+2}...+c_{b-1}+c_{b}$
Products: $\prod_{i=a}^{b}c_i=c_a\cdot c_{a+1}\cdot c_{a+2}...\cdot c_{b-1}\cdot c_{b}$

$\sum_{i=a}^{b}f_1(i)+f_2(i)+\ldots f_n(i)=\sum_{i=a}^{b}f(i)+\sum_{i=a}^{b}f_2(i)+\ldots+\sum_{i=a}^{b}f_n(i)$

$\sum_{i=a}^{b}c\cdot f(i)=c\cdot \sum_{i=a}^{b}f(i)$

$\sum_{i=1}^{n} i= \frac{n(n+1)}{2}$ , and in general $\sum_{i=a}^{b} i= \frac{(b-a+1)(a+b)}{2}$

The above should all be self-evident and provable by the reader within seconds.

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

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

$\sum_{i=1}^n i^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$

$\sum_{i=1}^n i^5=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}$

$\prod_{i=a}^{b}x=x^{(b-a+1)}$

$\prod_{i=a}^{b}x\cdot y=x^{(b-a+1)}y^{(b-a+1)}$

These should be self-evident, as above.

@@ Line 31: / Line 31: @@
 <math>\prod_{i=a}^{b}x\cdot y=x^{(b-a+1)}y^{(b-a+1)}</math>
-<!-- same as above, there are others.... but no fancy ones like divergence/convergence, please -->
+These should be self-evident, as above.
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