User:Temperal/The Problem Solver's Resource3

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Revision as of 13:13, 3 January 2008 by Temperal (talk | contribs) (Rules of Summation: equal signs)



The Problem Solver's Resource
Introduction | Other Tips and Tricks | Methods of Proof | You are currently viewing page 3.

Summations and Products

Definitions

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

Rules of Summation

$\sum_{i=a}^{b}f(i)+g(i)=\sum_{i=a}^{b}f(i)+\sum_{i=a}^{b}g(i)$

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

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

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

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

$\sum_{i\equal{}1}^n i^4=\frac{n(n\plus{}1)(2n\plus{}1)(3n^2\plus{}3n\minus{}1)}{30}$ (Error compiling LaTeX. Unknown error_msg)

$\sum_{i\equal{}1}^n i^5=\frac{n^2(n\plus{}1)^2(2n^2\plus{}2n\minus{}1)}{12}$ (Error compiling LaTeX. Unknown error_msg)


Rules of Products

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

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


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