Better Students Ask More Questions.
This is the other part of question...
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.
1 Answer | add yours
Best answer as selected by question asker.
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))`
Posted by jeew-m on June 19, 2013 at 6:32 PM (Answer #1)
Join to answer this question
Join a community of thousands of dedicated teachers and students.