Square of odd numbers can be expressed as:

\frac{n^{2} - 1}{2} + \frac{n^{2} + 1}{2}

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\frac{n^{2} - 3}{2} + \frac{n^{2} + 3}{2}

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n^{2} - 1 + n^{2} + 1

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n^{2} + 1 + n^{2} + 1

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Solution

The correct option is A $\frac{n^{2} - 1}{2} + \frac{n^{2} + 1}{2}$
Square of any odd number can be expressed as a sum of 2 consecutive natural numbers.
For example
$3^{2} = 9$
9 can also be written as
$9 = 4 + 5$
$9 = \frac{3^{2} - 1}{2} + \frac{3^{2} + 1}{2}$

Hence,
$n^{2} = \frac{n^{2} - 1}{2} + \frac{n^{2} + 1}{2}$

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Similar questions

n^{2} = \frac{n^{2} - 1}{2} + \frac{n^{2} + 1}{2}

If $n$ is odd, then $1 + 3 + 5 + 7 + . . . . . +$ to $n$ terms is equal to

(a) $(n^{2} + 1)$

(b) $(n^{2} - 1)$

(d) $(2 n^{2} + 1)$

{lim}_{n \to \infty} (\frac{n}{n^{2}} + \frac{n}{n^{2} + 1} + \frac{n}{n^{2} + 2^{2}} + . . . . . . + \frac{n}{2 n^{2} - 2 n + 1})

{sec}^{- 1} \sqrt{2} + {sec}^{- 1} \frac{\sqrt{10}}{3} + {sec}^{- 1} \frac{\sqrt{50}}{7} + . . . + {sec}^{- 1} \sqrt{\frac{(n^{2} + 1) (n^{2} - 2 n + 2)}{(n^{2} - n + 1)^{2}}}

N_{2} + O_{2} ⇌ 2 N O, K_{1}

(\frac{1}{2}) N_{2} + (\frac{1}{2}) O_{2} ⇌ N O, K_{2}

2 N O ⇌ N_{2} + O_{2}; K_{3}

N O ⇌ (\frac{1}{2}) N_{2} + (\frac{1}{2}) O_{2}; K_{4}

K_{1}, K_{2}, K_{3}

K_{4}

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