Question

# Prove by the principle of mathematical induction that 1+2+3+...+n<(2n+1)28, for all n∈N.          Or Using the principle of mathematical induction, prove that 1.3+2.32+3.33+...+n.3n=(2n−1)3n+1+34. for all n∈N.

Solution

## Let P(n) be the statement given by P(n):1+2+3+...+n<(2n+1)28 For n=1, P(1):1<(2+1)28 ∵           1<98 ∴   P(1) is true. Assume that, P(k) be true, for some natural number k. 1+2+3+...+k <(2k+1)28 Now, we will show that P(k+1) is true whenever P(k) is true. i.e. 1+2+3+...+k+k+1<{2(k+1)+1}28 Now, P(k) is true. ⇒           1+2+3+...+k <(2k+1)28 ⇒           1+2+3+...+k+(k+1)<(2k+1)28+(k+1)                                                                        [adding both sides by (k+1)] ⇒           1+2+3+...+k+(k+1)<18{(2k+1)2+8(k+1)} ⇒           1+2+3+...+k+(k+1)<18(4k2+4k+1+8k+8) ⇒           1+2+3+...+k+(k+1)<18(4K2+12K+9) ⇒           1+2+3+...+K+(K+1)<18(2k+3)2 ⇒           1+2+3+...+k+(k+1)<18{2(k+1)+1}2 Thus, P(k+1) is true. Hence, by principle of mathematical induction, P(n) is true, ∀ n∈N. Let P(n) be the statement given by P(n):1.3+2.32+3.32+...+n.3n=(2n−1)3n+1+34 For          n=1:P(1):l.3={2(1)−1}31+1+34 ⇒           3=9+34 ⇒ 3=3 So, P(1) is true. Assume that P(k) be true, for some natural number k. ∴ 1.3+2.32+3.33+...+k.3k=(2k−1)3k+1+34        ...(i) Now, we will show that P(k+1) is true, whenever P(k) is true i.e. 1.3+2.32+3.33 +...+k.3k+(k+1).3k+1 = {2(k+1)−1}3(k+1)+1+34 Now, 1.3+2.32+3.32 +...+k.3k+(k+1).3k+1 = (2k−1)3k+1+34+(k+1).3k+1         [from Eq.  (i)] = (2k−1)3k+1+3+4(k+1).3k+14 = (2k−1+4k+4).3k+1+34 = (6k+3).3k+1+34 = (2k+1).3k+2+34 = {2(k+1)−1}.3(k+1)+1+34 ∴  P(k+1) is true. Hence, by principle of mathematical induction, P(n) is true, ∀ n∈N.

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