Which of the following can be the square of a natural number 'n'?

Sum of the square of first n natural numbers.

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Sum of the first n natural numbers.

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Sum of the first (n-1) natural numbers.

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Sum of the first n odd natural numbers.

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Solution

The correct option is D Sum of the first n odd natural numbers.
The sum of first n odd natural number will be the square of a natural number n.

For example, sum of the first 7 odd consecutive numbers is 49.
$1 + 3 + 5 + 7 + 9 + 11 + 13 = 7^{2} = 49$

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Similar questions

The remainder obtained when the sum of squares first n odd natural numbers is divided the sum of squares of first n even natural numbers is

n

\frac{n^{2} - 1}{6}

n

n^{2}

n

\frac{n}{3} (4 n^{2} - 1)

\frac{1}{6} (2 n^{3} + 3 n^{2} + n)

\frac{1}{6} n (n + 1) (2 n + 1)

\frac{1}{6} (2 n^{3} + 3 n^{2} + n)

The ratio of squares of first n natural numbers to square of sum of first n natural numbers is17:325. The value of n is:

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