1.

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^2This 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»

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


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