If $(2 n + r) r, n \in N, r \in N$ is expressed as the sum of $k$ consecutive odd natural numbers, then $k$ is equal to

r

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n

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r + 1

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n + 1

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Solution

The correct option is B $r$
Let the $k$ consecutive odd natural numbers be $2 m + 1, 2 m + 3, . . . ., 2 m + k$ where $m ϵ N$
$∴ S_{k} = 2 m k + k^{2}$
Given $(2 n + r) r = 2 m k + k^{2}$
$\Rightarrow 2 n r + r^{2} = 2 m k + k^{2}$
This holds true when $n = m$ & $r = k$
Hence, A is correct.

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