Precalculus Examples

$\sum_{k = 10}^{60} 2 k$

Step 1

Split the summation to make the starting value of $k$ equal to $1$ .

$\sum_{k = 10}^{60} 2 k = \sum_{k = 1}^{60} 2 k - \sum_{k = 1}^{9} 2 k$

Step 2

Evaluate $\sum_{k = 1}^{60} 2 k$ .

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Step 2.1

Factor $2$ out of the summation.

$(2) \sum_{k = 1}^{60} k$

Step 2.2

The formula for the summation of a polynomial with degree $1$ is:

$\sum_{k = 1}^{n} k = \frac{n (n + 1)}{2}$

Step 2.3

Substitute the values into the formula and make sure to multiply by the front term.

$(2) (\frac{60 (60 + 1)}{2})$

Step 2.4

Simplify.

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Step 2.4.1

Simplify the expression.

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Step 2.4.1.1

Add $60$ and $1$ .

$2 \frac{60 \cdot 61}{2}$

Step 2.4.1.2

Multiply $60$ by $61$ .

$2 (\frac{3660}{2})$

Step 2.4.2

Cancel the common factor of $2$ .

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Step 2.4.2.1

Cancel the common factor.

$2 (\frac{3660}{2})$

Step 2.4.2.2

Rewrite the expression.

$3660$

Step 3

Evaluate $\sum_{k = 1}^{9} 2 k$ .

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Step 3.1

Factor $2$ out of the summation.

$(2) \sum_{k = 1}^{9} k$

Step 3.2

The formula for the summation of a polynomial with degree $1$ is:

$\sum_{k = 1}^{n} k = \frac{n (n + 1)}{2}$

Step 3.3

Substitute the values into the formula and make sure to multiply by the front term.

$(2) (\frac{9 (9 + 1)}{2})$

Step 3.4

Simplify.

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Step 3.4.1

Simplify the expression.

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Step 3.4.1.1

Add $9$ and $1$ .

$2 \frac{9 \cdot 10}{2}$

Step 3.4.1.2

Multiply $9$ by $10$ .

$2 (\frac{90}{2})$

Step 3.4.2

Cancel the common factor of $2$ .

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Step 3.4.2.1

Cancel the common factor.

$2 (\frac{90}{2})$

Step 3.4.2.2

Rewrite the expression.

$90$

Step 4

Replace the summations with the values found.

$3660 - 90$

Step 5

Subtract $90$ from $3660$ .

$3570$