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If sum(r=1)^(n) T(r) = (n)/(8) (n + 1) (...

If `sum_(r=1)^(n) T_(r) = (n)/(8) (n + 1) (n +2) (n + 3)`, then find `sum__(r=1)^(n) (1)/(T_(r))`

Text Solution

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`:. T_(n) = S_(n) - S_(n - 1)`
`=underset(r = 1)overset(n)sum T_(r) - underset(r = 1)overset(n -1)sum T_(r) = (n (n + 1) (n + 2) (n + 3))/(8) - ((n - 1) n (n + 1) (n + 2))/(8) = (n(n + 1) (n + 2))/(8) [(n + 3) - (n - 1)]`
`T_(n) = (n (n + 1) (n + 2))/(8) (4) = (n(n + 1) (n + 2))/(2)`
`rArr (1)/(T_(n)) = (2)/(n (n + 1) (n + 2)) = ((n + 2) - n)/(n(n + 1) (n + 2)) = (1)/(n(n + 1)) - (1)/((n + 1)(n + 2))`...(i)
Let `V_(n) = (1)/(n (n + 1))`
`:. (1)/(T_(n)) = V_(n) - V_(n + 1)`
Putting `n = 1, 2, 3, ... n`
`rArr (1)/(T_(1)) + (1)/(T_(2)) + (1)/(T_(3)) + ... + (1)/(T_(n)) = (V_(1) - V_(n + 1)) rArr underset(r = 1)overset(n)sum (1)/(T_(r)) = (n^(2) + 3n)/(2(n + 1) (n + 2))`
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