Question

n^2 + 3n is always divisible by which number, provided n is an integer?

(a) 2

(b) 3

(c) 4

(d) 5

I had been asked this question in my homework.

The doubt is from The Principle of Mathematical Induction topic in chapter Principle of Mathematical Induction of Mathematics – Class 11

Answer 1

Correct option is (a) 2

Easy explanation: P(n) = n^2 + 3n

P(1) = 1 + 3

P(1) = 4

Let’s assume that P(k) is true and divisible by 4. Therefore, P(k) = k^2 + 3k can be written as 4c.

We need to check if P(k + 1) is divisible by 4

P(k+1) = (k + 1)^2 + 3(k + 1)

P(k+1) = k^2 + 1 + 2k + 3k + 3

P(k+1) = k^2 + 5k + 4

P(k+1) = (k^2 + 3k) + 2k + 4

P(k+1) = 4c + 2k + 4

P(k+1) = 4c + 2(k + 2)

Clearly the second part of the equation is not divisible by 4. However P(k) = 4c is divisible by 2 and

P(k + 1) is also divisible by 2. Therefore, 2 divides P(n).