19. If n is a positive integer, then n(n+1)(n+2)is

(a) odd whenever n is odd

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

(b) even whenever n is even

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

(d) divisible by 4 whenever n is even

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

Open in App

Solution

The correct option is D (d) divisible by 4 whenever n is even
Ans: (d)
Sol:
n(n+1(n+2) will be even, whether n is even or odd because either (n+1) or (n+2) will be even.

n(n+1)(n+2) will be divisible by 3, whether n is even or odd, because either of n or n+1 or n+2 will be a divisor of 3.

n(n+1) (n+2) will be divisible by 4 whenever n is even because either n or n+2 will be a divisor of 4.

Suggest Corrections

Similar questions

n (n + 1) (n + 2)

n

n (n + 1) (n + 2)

n

k

n

n \sum k = 1 k^{3} {(\frac{^{n} C_{k}}{^{n} C_{k - 1}})}^{2} = \frac{n {(n + 1)}^{2} (n + 2)}{12}

n

n^{n + 1} - n {(n - 1)}^{n + 1} + \frac{n (n - 1)}{2!} {(n - 2)}^{n + 1} - \dots = \frac{1}{2} n (n + 1)!

n^{n} - (n + 1) {(n - 1)}^{n} + \frac{(n + 1) n}{2!} {(n - 2)}^{n} - \dots = 1

n

1^{n} - n 2^{n} + \frac{n (n - 1)}{1 \cdot 2} 3^{n} - \dots = {(- 1)}^{n} n!

{(n + p)}^{n} - n {(n + p - 1)}^{n} + \frac{n (n - 1)}{2!} {(n + p - 2)}^{n} - \dots = n!

n + 1

n

n \sum r = 1 r^{3} {(\frac{^{n} C_{r}}{^{n} C_{r - 1}})}^{2} = \frac{n {(n + 1)}^{2} (n + 2)}{12}

Join BYJU'S Learning Program