Skip to main contentSkip to solution

What is the value of (11n)+(12n)+(13n)+\left(1-\frac{1}{n}\right)+\left(1-\frac{2}{n}\right)+\left(1-\frac{3}{n}\right)+\cdots upto n terms?

Solution

Correct Option: 4

Sum =n1n(1+2++n)=n1nn(n+1)2=nn+12=2nn12=n12= 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}.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question