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Question
Find the sum of the following.
`(1 - 1/n) +(1 -2/n) + (1- 3/n) +` ......up to n terms.
Solution
On simplifying the given series, we get:
`(1 - 1/n) +(1 -2/n) + (1- 3/n) +` .... n terms
`=(1+1+1+ .......... "n terms") - (1/n+2/n+3/n+ .......+n/n)`
`= n - ( 1/n + 2/n +3/n + ..................+ n/n)`
` Here, (1/n +2/n +3/n +........+n/n) " is an AP whose first term is" 1/2 "and the common difference"`
is `(2/n - 1/n) = 1/n.`
The sum of terms of an AP is given by
`s_n = n/2 [ 2a + (n-1) d ] `
`= n - [ n/2{ 2 xx (1/n) + (n-1) xx (1/n) }]`
` = n -[ n/2 [ (2/n) + ((n-1)/n) ]] = n- { n/2((n+1)/n) }`
`=n- ((n+1)/2) = (n-1)/2`
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