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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
\[\frac{1}{n}\]
\[\frac{n - 1}{n}\]
\[\frac{n + 1}{2n}\]
\[\frac{n + 1}{n}\]
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Solution
\[\frac{n + 1}{n}\]
Given:
Sum of the even natural numbers = k\[\times\] Sum of the odd natural numbers
\[\frac{n}{2}\left\{ 2a + \left( n - 1 \right)d \right\} = k \times \frac{n}{2}\left\{ 2a + \left( n - 1 \right)d \right\}\]
\[ \Rightarrow \left\{ 2 \times 2 + \left( n - 1 \right)2 \right\} = k \times \left\{ 2 \times 1 + \left( n - 1 \right)2 \right\}\]
\[ \Rightarrow \frac{4 + \left( n - 1 \right)2}{2 + \left( n - 1 \right)2} = k\]
\[ \Rightarrow \frac{n + 1}{n} = k\]
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