Question

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

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Answer 1

Answer:

sum of the squares of first n natural numbers

$$\begin{lgathered}S_{n} = 1^{2} + 2^{2} + 3^{2} + \cdot\cdot \cdot + n^{2} \\= \frac{n(n+1)(2n+1)}{6}\end{lgathered}$$

(ii) sum of the first n natural numbers

$$\begin{lgathered}S_{n} = 1+2+3+\cdot\cdot\cdot + n \\= \frac{n(n+1)}{2}\end{lgathered}$$

(iii) sum of first (n – 1) natural numbers

$$\begin{lgathered}S_{n} = 1 + 2 + 3 + \cdot \cdot \cdot + (n-1) \\= \frac{(n-1)(n-1+1)}{2} \\= \frac{(n-1)n}{2}\end{lgathered}$$

(iv) sum of first ‘n’ odd natural numbers.

$$\begin{lgathered}\pink { S_{n}= 1+3+5+\cdot \cdot \cdot + n\:terms }\\ \green { = n^{2}}\end{lgathered}$$

Therefore.,

$$Option \: \pink { (iv) } \: is \: correct .$$

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Answer 2

Answer:

Sum of the square of first n natural number.