If $k_{1}, k_{2}, k_{3}, . . ., k_{n}$ are odd natural numbers, then the remainder when $k_{1}^{2} + k_{2}^{2} + k_{3}^{2} + . . . + k_{n}^{2}$ is divided by 4, is always equal to

\frac{n^{2}}{4}

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\frac{n}{4}

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\frac{n}{2}

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\frac{n + 1}{2}

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Solution

As $k_{1}, k_{2}, k_{3}, . . ., k_{n}$ are odd natural numbers, we have
$k_{1} = 2 m_{1} + 1$
$k_{2} = 2 m_{2} + 1$
$k_{3} = 2 m_{3} + 1$
⋮
$k_{n} = 2 m_{n} + 1$
where $m_{1}, m_{2}, m_{3}, . . ., m_{n}$ are natural numbers.
Now, $k_{1}^{2} + k_{2}^{2} + . . . + k_{n}^{2}$
$= (2 m_{1} + 1)^{2} + (2 m_{2} + 1)^{2} + . . . + (2 m_{n} + 1)^{2}$
$= (4 m_{1}^{2} + 4 m_{1} + 1) + (4 m_{2}^{2} + 4 m_{2} + 1) + . . . + (4 m_{n}^{2} + 4 m_{n} + 1)$
$= 4 (m_{1}^{2} + m_{2}^{2} + . . . + m_{n}^{2}) + 4 (m_{1} + m_{2} + . . . + m_{n}) + (1 + 1 + . . . + 1)$
$= 4 α + 4 β + n$
$(w h e r e α = m_{1}^{2} + m_{2}^{2} + . . . + m_{n}^{2} a n d β = m_{1} + m_{2} + . . . + m_{n})$
=4(α+β)+n
$T h u s, k_{1}^{2} + k_{2}^{2} + . . . + k_{n}^{2}$ when divided by 4 gives remainder $n .$
Hence, the correct answer is option (b).

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