Sums of Natural Numbers
What are Series
Series
- A series is a sum of a finite or infinite number of terms
- \(\sum\) is the notation used to define a series
- E.G. \(\(\sum_{r=1}^{5} (2r+1)\)\)
- The answer to this would be \(3+5+7+9+11 = 35\)
Essential Formulae
The formulae
- n: \(\(\sum_{r=1}^{n}1=n\)\)
- Sum of the first n integers: \(\(\sum_{r=1}^{n}r=1+2+3+\dots+n = \frac{n\times (n+1)}{2}\)\)
- Sum of the first n squares: \(\(\sum_{r=1}^{n}r^{2}=\frac{1}{6}n(n+1)(2n+1)\)\)
- Sum of the first n cubes: \(\(\sum_{r=1}^{n}r^{3}=\frac{1}{4}n^{2}(n+1)^2\)\)
- Subtracting series: \(\(\sum_{r=a}^nf(r)=\sum_{r=1}^nf(r)-\sum_{r=1}^{a-1}f(r)\)\)
- Variables: \(\(\sum_{r=1}^nkf(r)=k\sum_{r=1}^nf(r)\)\)
Subtracting Series
Subtracting series
- To add up all the integers from 3 to 9: \(\(\sum_{n=1}^9r-\sum_{r=1}^2r\)\)
- This gives us \(\(\sum_{n=3}^9r\)\)
With algebra
- Example with algebra: \(\(\sum_{r=n}^{3n}r=\sum_{r=1}^{3n}r-\sum_{r=1}^{n-1}r\)\)
-
\[=\frac{1}{2}(3n)(3n+1)-\frac{1}{2}(n-1)(n)\]
-
\[=\frac{1}{2}n(3(3n+1)-(n-1))\]
-
\[=\frac{1}{2}n(9n+3-n+1)\]
-
\[\frac{1}{2}n(8n+4)\]
-
\[2n(2n+1)\]
Breaking Up Summations
Breaking up summations
-
\[\sum_{i=1}^n(a_{i}+b{i})=a_{1}+b_{1}+a_{2}+b_{2}+\dots+a_{n}+b_{n}\]
-
\[=a_{1}+a_{2}+a_{3}+a_{4}+\dots+a_{n}+b_{1}+b_{2}+b_{3}+b_{4}+\dots+b_{n}\]
-
\[=\sum_{i=1}^na_{i}+\sum_{i=1}^nb_{i}\]
Show that \(\sum_{r=1}^n(3r+2)=\frac{n}{2}(3n+7)\)
-
\[\sum_{r=1}^n3r+\sum_{r=1}^n2\]
-
\[\frac{3}{2}n(n+1)+2n\]
-
\[\frac{n}{2}[3(n+1)+4]\]
-
\[\frac{n}{2}(3n+3+4)\]
-
\[\frac{n}{2}(3n+7)\]