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

 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 $\sum_{i=a}^{b}f_1(i)+f_2(i)+\ldots f_n(i)=\sum_{i=a}^{b}f(i)+\sum_{i=a}^{b}f_2(i)+\ldots+\sum_{i=a}^{b}f_n(i)$ $\sum_{i=a}^{b}c\cdot f(i)=c\cdot \sum_{i=a}^{b}f(i)$ $\sum_{i=1}^{n} i= \frac{n(n+1)}{2}$, 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. $\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$ $\sum_{i=1}^{n} i^3 = \left(\sum_{i=1}^{n} i\right)^2 = \left(\frac{n(n+1)}{2}\right)^2$ $\sum_{i=1}^n i^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$ $\sum_{i=1}^n i^5=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}$

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

These should be self-evident, as above.