Partial Sums of Series

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Pre-Calculus › Partial Sums of Series

Questions 1 - 4

For the sequence

$\cdot$

Determine $\sum_{i=1}^{6}a_i$ .

Explanation

$\sum_{i=1}^{6}a_i$ is defined as the sum of the terms from to

Therefore, to get the solution we must add all the entries from from to as follows.

Simplify the sum.

$\small 1+3+5+...+(2n-1)$

$\small n^2$

$\small (n-1)^2$

$\small \frac{n(n+1)}{2}$

$\small \small \frac{(n-1)(n)}{2}$

$\small (n+1)^2$

Explanation

The answer is $\small n^2$ . Try this for $\small n=1,2,3,4$ :

$\small 1=1^2$

$\small 1+3=4=2^2$

$\small 1+3+5=9=3^2$

$\small 1+3+5+7=16=4^2$

This can be proven more generally using a proof technique called mathematical induction, which you will most likely not learn in high school.

In case you are not familiar with summation notation, note that: $\sum_{i=1}^{3}(\frac{1}{2})^{i}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}$

What is the value of $\sum_{i=0}^{3}\left(\frac{1}{2}\right)^{i}$ ?

$\frac{15}{8}$

$\frac{5}{8}$

$\frac{31}{16}$

Explanation

Because the iterator starts at , we first have a $\frac{1}{2}^0 = 1$ .

Now expanding the summation to show the step by step process involved in answering the question we get,

$\ \sum_{i=0}^{3}\left(\frac{1}{2}\right)^{i}=\left(\frac{1}{2}\right)^0+\left(\frac{1}{2}\right)^1+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3\ \=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\ \=\frac{8}{8}+\frac{4}{8}+\frac{2}{8}+\frac{1}{8}\ \=\frac{15}{8}$