Partial Sums of Series - Pre-Calculus

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For the sequence

$\cdot$

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

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$\sum_${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.

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}$$\frac{1}{2}$$ ?

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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$

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}$$\left($\frac{1}{2}\right)^{i}$$ ?

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Because the iterator starts at , we first have a .

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

Simplify the sum.

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

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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.

For the sequence

$\cdot$

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

Tap to see back →

$\sum_${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.

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}$$\frac{1}{2}$$ ?

Tap to see back →

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$

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}$$\left($\frac{1}{2}\right)^{i}$$ ?

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Because the iterator starts at , we first have a .

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

Simplify the sum.

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

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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.

For the sequence

$\cdot$

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

Tap to see back →

$\sum_${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.

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}$$\frac{1}{2}$$ ?

Tap to see back →

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$

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}$$\left($\frac{1}{2}\right)^{i}$$ ?

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Because the iterator starts at , we first have a .

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

Simplify the sum.

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

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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.

For the sequence

$\cdot$

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

Tap to see back →

$\sum_${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.

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}$$\frac{1}{2}$$ ?

Tap to see back →

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$

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}$$\left($\frac{1}{2}\right)^{i}$$ ?

Tap to see back →

Because the iterator starts at , we first have a .

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

Simplify the sum.

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

Tap to see back →

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.