Non-perfect square numbers lying between $n^{2}$ & $(n + 1)^{2}$ are

2 n

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

n^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(n + 1)^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $2 n$
$(n + 1)^{2} - n^{2} = (n^{2} + 2 n + 1) - n^{2} = 2 n + 1$
And numbers lying is $1$ less than the difference.
So, $2 n + 1 - 1 = 2 n$
Therefore, option A is the correct answer.

Suggest Corrections

Similar questions

n^{2}

(n + 1)^{2}

2 n

n^{2}

(n + 2)^{2}

n^{2} + (n + 1)^{2} + {(n^{2} + n)}^{2} = {(n^{2} + n + 1)}^{2}

lim n \to \infty [\frac{n}{n^{2} + 1^{2}} + \frac{n}{n^{2} + 2^{2}} + . . . . . + \frac{1}{2 n}]

lim n \to \infty [\frac{1}{\sqrt{n^{2} - 1^{2}}} + \frac{1}{\sqrt{n^{2} - 2^{2}}} + \dots + \frac{1}{\sqrt{2 n - 1}}] =

Join BYJU'S Learning Program