Summation

Tags

Calculus

Cegep/2

Word count

258 words

Reading time

2 minutes

Properties

• $\sum _{i=1}^{n}k=kn$
• $\sum _{i=1}^{n}k{a}_i=k\sum _{i=1}^n{a}_i$
• $\sum _{i=1}^n(a_i+b_i)=\sum _{i=1}^n{a}_i+\sum _{i=1}^n{b}_i$
• $\sum _{i=1}^{n}i=\frac{n(n+1)}{2}$
• ==$\sum _{i=1}^n{i}^2=\frac{n(n+1)(2n+1)}{6}$==
• ==$\sum _{i=1}^n{i}^3={\left(\frac{n(n+1)}{2}\right)}^2$==

[!example]- Find the value of the sum as an expression of $n$: $\sum _{i=1}^n({(2+\frac{3i}{n})}^2-(2+\frac{3i}{n}))\left(\frac{3}{n}\right)$

$$\begin{array}{rl}\sum _{i=1}^n({(2+\frac{3i}{n})}^2-(2+\frac{3i}{n}))\left(\frac{3}{n}\right)& =\frac{3}{n}(\sum _{i=1}^n{(2+\frac{3i}{n})}^2-\sum _{i=1}^n(2+\frac{3i}{n}))\\ & =\frac{3}{n}(\sum _{i=1}^n(4+\frac{12i}{n}+\frac{9i^2}{n^2})-2n-\sum _{i=1}^n\frac{3i}{n})\\ & =\frac{3}{n}(4n+\frac{12}{n}\sum _{i=1}^{n}i+\frac{9}{n^2}\sum _{i=1}^n{i}^2-2n-\frac{3}{n}\sum _{i=1}^{n}i)\\ & =\frac{3}{n}(2n+\frac{9}{n}\cdot \frac{n(n+1)}{2}+\frac{9}{n^2}\cdot \frac{n(n+1)(2n+1)}{6})\\ & =6+\frac{27(n+1)}{2n}+\frac{9(n+1)(2n+1)}{2n^2}\end{array}$$

Contributors

Ajitani Hifumi

Changelog

Last edited about 2 hours ago

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