Question

If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers, then write the value of k.

Answer 1

It is given that the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers. $\therefore 2+4+\mathrm{6...2}n=k(1+3+\mathrm{5...}n)\Rightarrow 2(1+2+\mathrm{3...}n)=k(1+3+\mathrm{5...}n)\Rightarrow 2\times \frac{n(n+1)}{2}=k(1+3+\mathrm{5...}n)\Rightarrow \frac{n(n+1)}{2}=k\times n^2\because \text{sum of first odd natural numbers is}{n}^2\Rightarrow \frac{n(n+1)}{n^2}=kHence,k=\frac{n+1}{n}$