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

Revision as of 17:00, 5 January 2008

The Problem Solver's Resource

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

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

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

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

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

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

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

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

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

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

@@ Line 21: / Line 21: @@
 <math>\sum_{i=1}^{n} i^3 = \left(\sum_{i=1}^{n} i\right)^2 = \left(\frac{n(n+1)}{2}\right)^2</math>
-<math> \sum_{i\equal{}1}^n i^4=\frac{n(n\plus{}1)(2n\plus{}1)(3n^2\plus{}3n\minus{}1)}{30}</math>
+<math> \sum_{i\equal{}1}^n i^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}</math>
-<math> \sum_{i\equal{}1}^n i^5=\frac{n^2(n\plus{}1)^2(2n^2\plus{}2n\minus{}1)}{12}</math>
+<math> \sum_{i\equal{}1}^n i^5=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}</math>
 <!--there are others, I forgot what they were. Could someone please fill them in? -->