1.

n^2 + 3n is always divisible by which number, provided n is an integer?(a) 2(b) 3(c) 4(d) 5I 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»

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).


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