Number of integral values of n for which the quantity ${n + i}^{4}$ where $i^{2} = - 1$ , is an integer is

1

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

2

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

3

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

Infinite

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

Open in App

Solution

The correct option is D Infinite
Simply $n + i^{4}$ = $n + 1$ which is integer for every integral value of n .
So there are infinite values of n .

Suggest Corrections

Similar questions

1 + i^{2} + i^{4} + i^{6} + . . . + i^{2 n}

i = \sqrt{- 1}, n \in N

1, 2, 3, . . . ., 100

\frac{1^{2} + 2^{2} + 3^{3} + . . . . . . + n^{2}}{1 + 2 + 3 + . . . . . + n}

a = \sum_{n = 1}^{\infty} (\frac{2 n}{(2 n - 1)!})

b = \sum_{n = 1}^{\infty} (\frac{2 n}{(2 n + 1)!})

n \geq 1

\frac{n^{3} + 3}{n^{2} + 7}

x_{1} + x_{2} + x_{3} + x_{4} \leq n

Join BYJU'S Learning Program