Question

By induction hypothesis, the series 1^2 + 2^2 + 3^2 + … + p^2 can be proved equivalent to ____________

(a) \(\frac{p^2+2}{7}\)

(b) \(\frac{p*(p + 1)*(2p + 1)}{6}\)

(c) \(\frac{p*(p+1)}{4}\)

(d) p+p^2

This question was addressed to me in an interview for internship.

Origin of the question is Principle of Mathematical Induction topic in division Induction and Recursion of Discrete Mathematics

Answer 1

The correct answer is (b) \(\frac{p*(p + 1)*(2p + 1)}{6}\)

The best explanation: By principle of mathematical induction, we now assume that p (b) is true 1^2 + 2^2 + 3^2 + … + b^2 = \(\frac{b (b + 1) (2b + 1)}{6}\)

so to prove P(b+1): 1^2 + 2^2 + 3^2 + … + b^2 + (b + 1)^2 = \(\frac{b (b + 1) (2b + 1)}{6}\) + (b + 1)^2

By induction assumption it is shown that 1^2 + 2^2 + 3^2 + … + b^2 + (b + 1)^2 = \(\frac{(b + 1) [(b + 2) (2b + 3)]}{6}\). Hence it is proved that 1^2 + 2^2 + 3^2 + … + p^2 = \(\frac{p*(p + 1)*(2p + 1)}{6}\).