$n^{2} - n + 1$ is an odd number for all

n > 1

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

n > 2

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

n \geq 1

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

n \geq 5

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

Open in App

Solution

The correct option is C $n \geq 1$
For $n = 1$ , we have $n^{2} - n + 1 = 1^{2} - 1 + 1 = 1$ which is an odd number
For $n = 2$ , we have $n^{2} - n + 1 = 2^{2} - 2 + 1 = 3$ which is an odd number
For $n = 3$ , we have $n^{2} - n + 1 = 3^{2} - 3 + 1 = 7$ which is an odd number
For $n = 4$ , we have $n^{2} - n + 1 = 4^{2} - 4 + 1 = 13$ which is an odd number
For $n = 5$ , we have $n^{2} - n + 1 = 5^{2} - 5 + 1 = 19$ which is an odd number
Hence, $n^{2} - n + 1 = 1^{2} - 1 + 1 = 1$ is an odd number for all $n \geq 1$

Suggest Corrections

Similar questions

n^{2} + n + 1

n \in N

n

(n, \frac{n^{2} - 1}{2}, \frac{n^{2} + 1}{2})

n

Prove that $1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + . . . . . + \frac{1}{n^{2}} < 2 - \frac{1}{2}$ for all $n > 2$ , n $ϵ$ N.

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

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

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

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

1^{n} + 2^{n} + 3^{n} + . . . . . + 15^{n}

480

n \geq 5

Join BYJU'S Learning Program