solute for 1/1*4+1/4*7+.....................+1/(3n-2)(3n+1)=...

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You need to calculate the sum `sum_(k=1)^(3n)` `1/((3k-2)(3k+1)).`

You need to use the partial fraction decomposition to write the fraction as an addition of two elementary fractions such that:

`1/((3k-2)(3k+1)) = A/(3k-2) + B/(3k+1)`

Multiplying all the equation by the common denominator` (3k-2)(3k+1)` yields:

`1 = A(3k+1) + B(3k-2)`

Opening the brackets yields:

`1 = 3kA + A + 3kB - 2B`

Equating the coefficients of like powers of k yields:

`3A + 3B= 0 =gt A+B = 0`

`A - 2B = 1 =gt -B - 2B = 1 =gtB = -1/3 =gt A = 1/3`

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

Plugging values for k in the equation above yields:

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

`1/((1)*(4)) = (1/3)*(1/3 - 1/4)`

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

`1/((4)*(7)) = (1/3)*(1/(4) - 1/(7))`

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

Adding the terms from left side and the terms from right side yields:

`sum_(k=1)^n``1/((3k-2)(3k+1)) = (1/3)*(1/3 - 1/4 + 1/4 - 1/7 + 1/7 - .....- 1/(3n-2)+ 1/(3n-2) - 1/(3n+1))`

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

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

**Evaluating the given sum by partial fraction decomposition technique yields `sum_(k=0)^n 1/((3k-2)(3k+1)) = (1/3)*( (3n - 2)/(3n + 1)).` **

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