InterviewSolution
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InterviewSolution
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Prove the following by the principle ofmathematical induction: ` 1. 3+2. 4+3. 5++(2n-1)(2n+1)=(n(4n^2+6n-1))/3` |
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Answer» Let the given statement be P(n). Then, `P(n): 1*3+3*5+5*7+…+(2n-1)(2n+1)= 1/3 n(4n^(2)+6n-1)`. When n = 1, we have LHS = ` 1*3 and RHS = 1/3 xx 1 xx(4xx1^(2)+6xx1-1)=1/3 xx 1 xx 9 = 3`. ` :. ` LHS = RHS. Thus, P(1) is true. Let P(k) be true. Then, `P(k): 1*3+3*5+5*7+...+(2K-1)(2k+1)= 1/3 k(4k^(2)+6k-1)." "` ...(i) Now, ` 1*3+3*5+5*7+...+(2k-1)(2k+1)+{2(k+1)-1}{2(k+1)+1}` ` ={1*3+3*5+5*7+...+(2k-1)(2k+1)}+(2k+1)(2k+3)` ` = 1/3 k(4k^(2)+6k-1)+(2k+1)(2k+3)" "`[using (i)] ` = 1/3 [(4k^(3)+6k^(2)-k)+3(4k^(2)+8k+3)] = 1/3 (4k^(3)+18k^(2)+23k+9)` ` = 1/3 (k+1)(4k^(2)+14k+9)= 1/3 (k+1)[4(k+1)^(2)+6(k+1)-1]`. ` :. P(k+1): 1*3+3*5+5*7+...+{2(k+1)-1}{2(k+1)+1}` ` = 1/3 (k+1){4(k+1)^(2)+6(k+1)-1}`. Thus, P(k+1) is true, whenever P(k) is true. `:. ` P(1) is true and P(k+1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction , we have ` 1*3+3*5+5*7+...+(2n-1)(2n+1) = 1/3 n (4n^(2)+6n-1)" for all " n in N`. |
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