Question

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

Options

• \[\frac{1}{n}\]

• \[\frac{n - 1}{n}\]

• \[\frac{n + 1}{2n}\]

• \[\frac{n + 1}{n}\]

Answer 1

\[\frac{n + 1}{n}\]

Given:
Sum of the even natural numbers = k\[\times\] Sum of the odd natural numbers

\[\frac{n}{2}\left\{ 2a + \left( n - 1 \right)d \right\} = k \times \frac{n}{2}\left\{ 2a + \left( n - 1 \right)d \right\}\]

\[ \Rightarrow \left\{ 2 \times 2 + \left( n - 1 \right)2 \right\} = k \times \left\{ 2 \times 1 + \left( n - 1 \right)2 \right\}\]

\[ \Rightarrow \frac{4 + \left( n - 1 \right)2}{2 + \left( n - 1 \right)2} = k\]

\[ \Rightarrow \frac{n + 1}{n} = k\]