Q12Multiple choice
= 1 = . Again for some k ≥ 1, k2 = . Now we 6 6 prove that (k + 1) ((k + 1) + 1) (2(k + 1) + 1) (k + 1)2 = Solution False Since in the inductive step both the inductive hypothesis and what is to be proved are wrong.
= 1 = . Again for some k ≥ 1, k2 = . Now we 6 6 prove that (k + 1) ((k + 1) + 1) (2(k + 1) + 1) (k + 1)2 = Solution False Since in the inductive step both the inductive hypothesis and what is to be proved are wrong.