# Using method of differences, find the sum of the first n terms of the series. `sum_(n=1)^n(2n-1)/(n(n+1)(n+2))`

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You need to split the fraction into three simpler fractions (there exists 3 factors to denominator), hence, you need to use partial fraction decomposition, such that:

`(2n - 1)/((n(n+1)(n+2))) = a/n + b/(n+1) + c/(n+2)`

`2n - 1 = a(n+1)(n+2) + b*n(n+1) + c*n(n + 1)`

`2n - 1 = an^2 + 3an + 2a + bn^2 + bn + cn^2 + cn`

`2n - 1 = n^2(a + b) + n(3a + b + c) + 2a`

Equating the coefficients of like powers, yields:

`{(a + b = 0),(3a + b + c = 2),(2a = -1):}`

`{(a = -b),(3a + b + c = 2),(a = -1/2):}`

`{(b = 1/2),(-3/2 + 1/2 + c = 2),(a = -1/2):}`

`{(b = 1/2),(c - 1 = 2),(a = -1/2):}`

`{(b = 1/2),(c = 3),(a = -1/2):}`

`(2n - 1)/((n(n+1)(n+2))) = -1/(2n) + 1/(2(n+1)) + 3/(n+2)`

You need to evaluate the given summation, hence, you may replace `-1/(2n) + 1/(2(n+1)) + 3/(n+2)` for the fraction `(2n - 1)/((n(n+1)(n+2)))` , such that:

`sum_(k = 1)^n (2k - 1)/((k(k+1)(k+2))) = sum_(k = 1)^n (-1/(2k) + 1/(2(k+1)) + 3/(k+2))`

Giving values to k, yields:

`k = 1 => -1/2 + 1/4 + 3/3`

`k = 2 => -1/4 + 1/6 + 3/4`

`k = 3 => -1/6 + 1/8 + 3/5`

`k = 4 => -1/8 + 1/10 + 3/6`

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`k = n => -1/2n + 1/(2(n+1)) + 3/(n+2)`

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`sum_(k = 1)^n (2k - 1)/((k(k+1)(k+2))) = -1/2 +1/(2(n+1)) + 3/3 + 3/4 + .. + 3/(n+2)`

`sum_(k = 1)^n (2k - 1)/((k(k+1)(k+2))) = -n/(2(n+1)) + 3 sum_(k = 0)^(n-1) 1/(k + 3)`

**Hence, evaluating the given summation, using the partial fraction expansion, yields **`sum_(k = 1)^n (2k - 1)/((k(k+1)(k+2))) = -n/(2(n+1)) + 3 sum_(k = 0)^(n-1) 1/(k + 3).`