CAS header

 

Find this content useful?
Think it should stay ad-free?

Summation Notation

Summation Notation

The use of the summation symbol allows us to quickly express the sum of items in a sequence. For example, the sum of the terms $2,5,8,11,\dots ,35$  ( a sequence of 12 terms) can be expressed as 

$$2 + 5 + 8 + 11 + \dots + 35 = \sum_{k=1}^{12} \left(3k-1\right) $$

The lower and upper indices describe the values that the variable (in this case $k$) will run over. So in this case, $k$ will go from $1$ to $12$. The symbol tells us to add up each term of the form $3k-1$ for each value of $k$ in this range.

The operation of summation obeys a number of useful rules

  1. Constants Law $$\sum_{k=1}^n \left(f(k) + c\right) = \left(\sum_{k=1}^n f(k)\right) + nc$$
  2. Linearity Part 1 $$\sum_{k=1}^n af(k) = a \sum_{k=1}^nf(k)$$
  3. Linearity Part 2 $$\sum_{k=1}^n \left(f(k)+ g(k)\right) = \sum_{k=1}^nf(k) + \sum_{k=1}^ng(k)$$
  4. Partial Sums $$\sum_{k=p}^n f(k) = \sum_{k=1}^n f(k) - \sum_{k=1}^{p-1}f(k)$$

We can apply these laws to our example above to decompose it into a simpler mathematical expression: $$ \sum_{k=1}^{12} \left(3k-1\right) = \sum_{k=1}^{12} \left(3k\right) - 12\times 1 = 3\sum_{k=1}^{12} \left(k\right) -12 $$ The expression $\sum_{k=1}^n k$ gives the $n^{th}$ triangular number, that is $$ \sum_{k=1}^n k = \frac{n(n+1)}{2} $$ Consequently, we get $$ \sum_{k=1}^{12} \left(3k-1\right) = 3\frac{12(13)}{2} - 12 = 222 $$

Summation Notation 1

Evaluate the expression $$\sum_{k=1}^{10}(20-2k)$$
solution - press button to display
$$\sum_{k=1}^{10}(20-2k)= 20\times 10 - 2\sum_{k=1}^{10}k = 200 - 2\frac{(10)(11)}{2} = 90$$

Summation Notation 2

Evaluate the expression $$\sum_{k=1}^{\infty}3\times \left(\frac{1}{2}\right)^k$$
solution - press button to display
Note that this summation is the sum to infinity of a geometric sequence! Nevertheless, we can use the laws of summations to somewhat simplify it $$ \sum_{k=1}^{\infty}3\times \left(\frac{1}{2}\right)^k = 3\sum_{k=1}^{\infty}\left(\frac{1}{2}\right)^k = 3\frac{\frac{1}{2}}{1-\frac{1}{2}}=3 $$