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

Revision as of 17:44, 29 September 2007

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

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

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

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

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

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

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

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

@@ Line 11: / Line 11: @@
 ===Rules of Summation===
-<math>\sum_{i=a}^{b}x+y=\sum_{i=a}^{b}x+\sum_{i=a}^{b}y</math>
+<math>\sum_{i=a}^{b}f(i)+g(i)=\sum_{i=a}^{b}f(i)+\sum_{i=a}^{b}g(i)</math>
-<math>\sum_{i=a}^{b}x\cdot y=x\cdot \sum_{i=a}^{b}y=y\cdot \sum_{i=a}^{b}x</math>
+<math>\sum_{i=a}^{b}c\cdot f(i)=c\cdot \sum_{i=a}^{b}f(i)</math>
+<math>\sum_{i=1}^{n} i= \frac{n(n+1)}{2}</math>, and in general <math>\sum_{i=a}^{b} i= \frac{(b-a+1)(a+b)}{2}</math>
+<math>\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}</math>
+<math>\sum_{i=1}^{n} i^3 = \left(\sum_{i=1}^{n} i\right)^2 = \left(\frac{n(n+1)}{2}\right)^2</math>
 <!--there are others, I forgot what they were. Could someone please fill them in? -->
@@ Line 19: / Line 25: @@
 ===Rules of Products===
-<math>\prod_{i=a}^{b}x+y+....+z=(x+y+....z)^(b-a+1)</math>
+<math>\prod_{i=a}^{b}x=x^{(b-a+1)}</math>
-<math>\prod_{i=a}^{b}x\cdot y=x^(b-a+1)y^(b-a+1)</math>
+<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 -->