# This is the other part of question http://www.enotes.com/homework-help/let-u-n-1-n-2-n-1-n-1-2-n-1-any-positive-integer-n-440354 find `V_n` such that `1/U_n = V_n-V_(n+1)` for positive integers n.

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### 1 Answer

From the previous part we have proved that;

`U_n = 1/6n(n+1)(n+2)`

`1/U_n = 6/(n(n+1)(n+2))`

Now we use partial fractions.

`6/(n(n+1)(n+2)) = A/n+B/(n+1)+C/(n+2)`

`6 = A(n+1)(n+2)+Bn(n+2)+Cn(n+1)`

When n = 0 then it gives A = 3

When n = -1 then it gives B = -6

When n = -2 then it gives C = 3

`1/U_n = 3/n-6/(n+1)+3/(n+2)`

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

`1/U_n = [3/n-3/(n+1)]-[3/(n+1)-3/(n+2)]`

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

If `V_n = 3/(n(n+1)) rarr V_(n+1) = 3/((n+1)(n+2))`

`1/U_n = V_n-V_(n+1)`

** So the answer is** `V_n = 3/(n(n+1))`