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If the sum of first n even natural number is equal to k times the sum of first n odd natural numbers, then k =
Answer & Solution
Correct Answer:
Option
D
Sum of n even natural number = n(n+1)
and sum of n odd natural numbers = n2
$$\eqalign{ & \therefore n\left( {n + 1} \right) = k{n^2} \cr & \Rightarrow k = \frac{{n(n + 1)}}{{{n^2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{n + 1}}{n} \cr} $$
and sum of n odd natural numbers = n2
$$\eqalign{ & \therefore n\left( {n + 1} \right) = k{n^2} \cr & \Rightarrow k = \frac{{n(n + 1)}}{{{n^2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{n + 1}}{n} \cr} $$
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