Q2Short answer
3 n 2n Solution Let the given statement be P(n), i.e., 1 1 1 n +1 P(n) : 1− 2 . 1− 2 ... 1− 2 = , for all natural numbers, n ≥ 2 2 3 n 2n We observe that P (2) is true, since 1 1 4 −1 3 2 + 1 1− 2 =1 − = = = 2 4 4 4 2× 2 Assume that P(n) is true for some k ∈ N, i.e., 1 1 1 k +1 P(k) : 1 − 2 . 1 − 2 ... 1 − 2 = 2 3 k 2k Now, to prove that P (k + 1) is true, we have 1 1 1 1 1− 2 . 1 − 2 ... 1 − 2 . 1 − 2 3 k (k + 1) 2 k +1 1 k 2 + 2k (k +1) +1 = 1− = = 2k (k +1) 2 2k ( k + 1) 2(k +1) Thus, P (k + 1) is true, whenever P(k) is true. Hence, by the Principle of Mathematical Induction, P(n) is true for all natural numbers, n ≥ 2.