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Hence or otherwise show that`sum_(r=1)^n 1/U_r = 3/2-3/((n+1)(n+2))`

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From the previous part I have proven that;

`1/U_n = V_n-V_(n+1) where V_n = 3/(n(n+1))`

`r = 1 rarr 1/U_1 = V_1-V_2`

`r = 2 rarr 1/U_2 = V_2-V_3`

`r = 3 rarr 1/U_3 = V_3-V_4`

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`r = (n-2) rarr 1/U_(n-2) = V_(n-2)-V_(n-1)`

`r = (n-1) rarr 1/U_(n-1) = V_(n-1)-V_(n)`

`r = n rarr 1/U_(n-2) = V_(n)-V_(n)+1`

If you add up the whole from r = 1 to r = n;

`sum_(r = 1)^n)1/U_r = V_1-V_(n+1)`

I think you have noticed how the values get cancel out. `V_2` at r = 1 will cancel out by `V_2` at r = 2.

`V_1 = 3/(1xx2) = 3/2`

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

*So the answer is;*

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

This is what you have asked to prove.

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