What is the value of \(\left(1-\frac{1}{n}\right)+\left(1-\frac{2}{n}\right)+\left(1-\frac{3}{n}\right)+\cdots\) up to \(n\) terms?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\frac{n-1}{2}\)
Each of the \(n\) terms is 1 minus a fraction, so the sum splits into \(n\) ones minus the sum of the fractions \(\frac{1}{n},\frac{2}{n},\ldots,\frac{n}{n}\), whose numerators are the first \(n\) natural numbers.
Sum of the ones \(=n\)
Sum of the fractions \(\displaystyle=\frac{1+2+\cdots+n}{n}=\frac{n(n+1)}{2n}=\frac{n+1}{2}\)
Required sum \(\displaystyle=n-\frac{n+1}{2}=\frac{2n-n-1}{2}=\frac{n-1}{2}\)
Tip: Options with \(n\) in them can be tested with a small value. For \(n=3\) the sum is \(\frac{2}{3}+\frac{1}{3}+0=1\); the options give \(\frac{1}{3}\), \(\frac{3}{2}\), 3 and 1, so only \(\frac{n-1}{2}\) fits.