Precalculus Examples

$\sum_{n = 51}^{100} 8 n$

Step 1

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

$\sum_{n = 51}^{100} 8 n = \sum_{n = 1}^{100} 8 n - \sum_{n = 1}^{50} 8 n$

Step 2

Evaluate $\sum_{n = 1}^{100} 8 n$ .

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Step 2.1

Factor $8$ out of the summation.

$(8) \sum_{n = 1}^{100} n$

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.

$(8) (\frac{100 (100 + 1)}{2})$

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 $100$ and $1$ .

$8 \frac{100 \cdot 101}{2}$

Step 2.4.1.2

Multiply $100$ by $101$ .

$8 (\frac{10100}{2})$

Step 2.4.2

Cancel the common factor of $2$ .

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Step 2.4.2.1

Factor $2$ out of $8$ .

$2 (4) \frac{10100}{2}$

Step 2.4.2.2

Cancel the common factor.

$2 \cdot 4 \frac{10100}{2}$

Step 2.4.2.3

Rewrite the expression.

$4 \cdot 10100$

Step 2.4.3

Multiply $4$ by $10100$ .

$40400$

Step 3

Evaluate $\sum_{n = 1}^{50} 8 n$ .

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Step 3.1

Factor $8$ out of the summation.

$(8) \sum_{n = 1}^{50} n$

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.

$(8) (\frac{50 (50 + 1)}{2})$

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 $50$ and $1$ .

$8 \frac{50 \cdot 51}{2}$

Step 3.4.1.2

Multiply $50$ by $51$ .

$8 (\frac{2550}{2})$

Step 3.4.2

Cancel the common factor of $2$ .

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Step 3.4.2.1

Factor $2$ out of $8$ .

$2 (4) \frac{2550}{2}$

Step 3.4.2.2

Cancel the common factor.

$2 \cdot 4 \frac{2550}{2}$

Step 3.4.2.3

Rewrite the expression.

$4 \cdot 2550$

Step 3.4.3

Multiply $4$ by $2550$ .

$10200$

Step 4

Replace the summations with the values found.

$40400 - 10200$

Step 5

Subtract $10200$ from $40400$ .

$30200$