Partial Sums of Series - Pre-Calculus

Card 0 of 16

Question

For the sequence

$\cdot$

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

Answer

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

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Question

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

Given the series above, what is the value of $\sum_{i=1}^{6}\frac{1}{2}$ ?

Answer

Since the upper bound of the iterator is and the initial value is , we need add one-half, the summand, six times.

This results in the following arithmetic.

$\ \sum_{i=1}^{6}\frac{1}{2}\ \=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\ \= \frac{6}{2}\ \=3$

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Question

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 $\sum_{i=0}^{3}\left(\frac{1}{2}\right)^{i}$ ?

Answer

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,

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

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Question

Simplify the sum.

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

Answer

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.

Compare your answer with the correct one above

Question

For the sequence

$\cdot$

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

Answer

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

Compare your answer with the correct one above

Question

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

Given the series above, what is the value of $\sum_{i=1}^{6}\frac{1}{2}$ ?

Answer

Since the upper bound of the iterator is and the initial value is , we need add one-half, the summand, six times.

This results in the following arithmetic.

$\ \sum_{i=1}^{6}\frac{1}{2}\ \=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\ \= \frac{6}{2}\ \=3$

Compare your answer with the correct one above

Question

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 $\sum_{i=0}^{3}\left(\frac{1}{2}\right)^{i}$ ?

Answer

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,

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

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Question

Simplify the sum.

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

Answer

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.

Compare your answer with the correct one above

Question

For the sequence

$\cdot$

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

Answer

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

Compare your answer with the correct one above

Question

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

Given the series above, what is the value of $\sum_{i=1}^{6}\frac{1}{2}$ ?

Answer

Since the upper bound of the iterator is and the initial value is , we need add one-half, the summand, six times.

This results in the following arithmetic.

$\ \sum_{i=1}^{6}\frac{1}{2}\ \=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\ \= \frac{6}{2}\ \=3$

Compare your answer with the correct one above

Question

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 $\sum_{i=0}^{3}\left(\frac{1}{2}\right)^{i}$ ?

Answer

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,

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

Compare your answer with the correct one above

Question

Simplify the sum.

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

Answer

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.

Compare your answer with the correct one above

Question

For the sequence

$\cdot$

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

Answer

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

Compare your answer with the correct one above

Question

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

Given the series above, what is the value of $\sum_{i=1}^{6}\frac{1}{2}$ ?

Answer

Since the upper bound of the iterator is and the initial value is , we need add one-half, the summand, six times.

This results in the following arithmetic.

$\ \sum_{i=1}^{6}\frac{1}{2}\ \=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\ \= \frac{6}{2}\ \=3$

Compare your answer with the correct one above

Question

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 $\sum_{i=0}^{3}\left(\frac{1}{2}\right)^{i}$ ?

Answer

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,

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

Compare your answer with the correct one above

Question

Simplify the sum.

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

Answer

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.

Compare your answer with the correct one above

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