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For any integer m>=3, the series 2+4+6+…+(4m) can be equivalent to ________(a) m^2+3(b) m+1(c) m^m(d) 3m^2+4The question was posed to me in class test.Question is taken from Principle of Mathematical Induction topic in division Induction and Recursion of Discrete Mathematics

Answer»

Right choice is (a) m^2+3

To EXPLAIN: The REQUIRED answer is m^2+3. Now, by induction assumption, we have to PROVE 2+4+6+…+4(k+1) = (k+1)^2+3also can be TRUE, 2+4+6+…+4(k+1) = 2+4+6+⋯+(4k+4) and by the subsequent steps, we can prove that (m+1)^2+3 also holds for m=k. So, it is proved.


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