QuestionCalculate the sum: $\sum_{n=2}^{6} \frac{n-1}{n(n+1)}$ .

Studdy Solution

STEP 1

Assumptions1. We are given a finite series that starts at n= and ends at n=6. . The general term of the series is $\frac{n-1}{n(n+1)}$ .

STEP 2

We need to calculate the sum of the series. The sum of a series is calculated by adding up all the terms of the series.
$\sum_{n=2}^{6} \frac{n-1}{n(n+1)} = \frac{2-1}{2(2+1)} + \frac{-1}{(+1)} + \frac{4-1}{4(4+1)} + \frac{5-1}{5(5+1)} + \frac{6-1}{6(6+1)}$

STEP 3

implify each term in the series.
$\sum_{n=2}^{6} \frac{n-1}{n(n+1)} = \frac{1}{6} + \frac{2}{12} + \frac{3}{20} + \frac{}{30} + \frac{5}{42}$

STEP 4

Further simplify each term in the series.
$\sum_{n=2}^{6} \frac{n-1}{n(n+1)} = \frac{1}{6} + \frac{1}{6} + \frac{3}{20} + \frac{2}{15} + \frac{}{42}$

STEP 5

Calculate the sum of the series by adding all the terms together.
$\sum_{n=2}^{} \frac{n-1}{n(n+1)} = \frac{1}{} + \frac{1}{} + \frac{3}{20} + \frac{2}{15} + \frac{5}{42} = \frac{1}{3} + \frac{3}{20} + \frac{2}{15} + \frac{5}{42}$

STEP 6

To add these fractions, we need to find a common denominator. The least common multiple of3,20,15, and42 is420.
$\frac{1}{3} + \frac{3}{20} + \frac{2}{15} + \frac{5}{42} = \frac{140}{420} + \frac{63}{420} + \frac{56}{420} + \frac{50}{420}$

STEP 7

Add the fractions together.
$\frac{140}{420} + \frac{63}{420} + \frac{56}{420} + \frac{50}{420} = \frac{309}{420}$

STEP 8

implify the fraction to its lowest terms.
$\frac{309}{420} = \frac{103}{140}$ The sum of the series $\sum_{n=2}^{6} \frac{n-1}{n(n+1)}$ is $\frac{103}{140}$ .

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