What is the value of (1-1/n)+(1-2/n)+(1-3/n)+*s upto n terms? Option 1: 1/n Option 2: 2/n Option 3: (n(n-1))/2 Option 4: (n-1)/2

Answer: (n-1)/2. Solution: Sum = n - 1/n(1 + 2 + s + n) = n - 1/n (n(n+1))/2 = n - (n+1)/2 = (2n - n - 1)/2 = (n-1)/2.

Quantitative Reasoning

Miscellaneous

Medium

What is the value of $\left(1-\frac{1}{n}\right)+\left(1-\frac{2}{n}\right)+\left(1-\frac{3}{n}\right)+\cdots$ upto n terms?

$\frac{1}{n}$

$\frac{2}{n}$

$\frac{n(n-1)}{2}$

$\frac{n-1}{2}$

✅ Correct Option: 4

Sum $= n - \frac{1}{n}(1 + 2 + \cdots + n) = n - \frac{1}{n} \cdot \frac{n(n+1)}{2} = n - \frac{n+1}{2} = \frac{2n - n - 1}{2} = \frac{n-1}{2}$ .

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