Partial Sums of Series - Pre-Calculus

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Question

For the sequence

$\cdot$

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

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

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

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Answer

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

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Question

Simplify the sum.

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

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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.

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Question

For the sequence

$\cdot$

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

Tap to reveal answer

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.

← Didn't Know|Knew It →

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

Tap to reveal answer

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

Tap to reveal answer

Answer

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

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Question

Simplify the sum.

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

Tap to reveal answer

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.

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Question

For the sequence

$\cdot$

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

Tap to reveal answer

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.

← Didn't Know|Knew It →

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

Tap to reveal answer

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

Tap to reveal answer

Answer

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

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Question

Simplify the sum.

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

Tap to reveal answer

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.

← Didn't Know|Knew It →

Question

For the sequence

$\cdot$

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

Tap to reveal answer

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.

← Didn't Know|Knew It →

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

Tap to reveal answer

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

Tap to reveal answer

Answer

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

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Question

Simplify the sum.

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

Tap to reveal answer

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.

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