Precalculus Examples

$\sum_{n = 0}^{50} 50 - n$

Step 1

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

$\sum_{n = 0}^{50} 50 - n = \sum_{n = 1}^{50} 50 - n + \sum_{n = 0}^{0} 50 - n$

Step 2

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

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Step 2.1

Split the summation into smaller summations that fit the summation rules.

$\sum_{n = 1}^{50} 50 - n = \sum_{n = 1}^{50} 50 - 1 \sum_{n = 1}^{50} n$

Step 2.2

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

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Step 2.2.1

The formula for the summation of a constant is:

$\sum_{k = 1}^{n} c = c n$

Step 2.2.2

Substitute the values into the formula.

$(50) (50)$

Step 2.2.3

Multiply $50$ by $50$ .

$2500$

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Simplify.

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Step 2.3.3.1

Add $50$ and $1$ .

$- 1 \frac{50 \cdot 51}{2}$

Step 2.3.3.2

Multiply $50$ by $51$ .

$- 1 (\frac{2550}{2})$

Step 2.3.3.3

Divide $2550$ by $2$ .

$- 1 \cdot 1275$

Step 2.3.3.4

Multiply $- 1$ by $1275$ .

$- 1275$

Step 2.4

Add the results of the summations.

$2500 - 1275$

Step 2.5

Subtract $1275$ from $2500$ .

$1225$

Step 3

Evaluate $\sum_{n = 0}^{0} 50 - n$ .

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Step 3.1

Expand the series for each value of $n$ .

$50 + 0$

Step 3.2

Add $50$ and $0$ .

$50$

Step 4

Replace the summations with the values found.

$1225 + 50$

Step 5

Add $1225$ and $50$ .

$1275$