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

(<span style="font-size:20px; color: blue;">Summations and Products</span>)
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===Rules of Summation===
 
===Rules of Summation===
<math>\sum_{i=a}^{b}x+y=\sum_{i=a}^{b}x+\sum_{i=a}^{b}y</math>
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<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>
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 +
<math>\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}</math>
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 +
<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? -->
 
<!--there are others, I forgot what they were. Could someone please fill them in? -->
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===Rules of Products===
 
===Rules of Products===
  
<math>\prod_{i=a}^{b}x+y+....+z=(x+y+....z)^(b-a+1)</math>
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<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 -->
 
<!-- same as above, there are others.... but no fancy ones like divergence/convergence, please -->

Revision as of 18:44, 29 September 2007



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

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 $\sum_{i=a}^{b} i= \frac{(b-a+1)(a+b)}{2}$

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

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


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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