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InterviewSolution
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Prove the following by the principle ofmathematical induction: `1/(3. 5)+1/(5. 7)+1/(7. 9)+1/((2n+1)(2n+3))=n/(3(2n+3))` |
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Answer» Let the given statement be P(n). Then, ` P(n) : 1/(3*5)+1/(5*7) + 1/(7*9) +…+ 1/((2n+1)(2n+3)) = n/(3(2n+3))`. Putting n = 1 in the given statement, we get LHS ` = 1/(3*5) = 1/15 and RHS = 1/(3(2xx1+3))= 1/15`. ` :. ` LHS = RHS. Thus, P(1) is true. Let P(k) be true. Then, `P(k): 1/(3*5)+1/(5*7)+1/(7*9) + ...+1/((2k+1)(2k+3)) = k/(3(2k+3)).`...(i) Now, ` 1/(3*5) + 1/(5*7) + ...+1/((2k+1)(2k+3))+1/({2(k+1)+1}{2(k+1)+3}) ` `{1/(3*5)+1/(5*7)+..+1/((2k+1)(2k+3))}+1/((2k+3)(2k+5))` ` = k/(3(2k+3))+1/((2k+3)(2k+5)) " "` [using (i)] `=(k(2k=5)+3)/(3(2k+3)(2k+5))= ((2k^(2)+5k+3))/(3(2k+3)(2k+5))=((k+1)(2k+3))/(3(2k+3)(2k+5)) ` `=((k+1))/(3(2k+5))= ((k+1))/(3{2(k+1)+3}).` `:. " " P(k+1) : 1/(3*5)+1/(5*7)+1/(7*9)+...+1/({2(k+1)+1}{2(k+1)+3}) = ((k+1))/(2{2(k+1)+3}).` This shows that 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*5)+1/(5*7)+1/(7*9)+...+ 1/((2n+1)(2n+3)) = n/(3(2n+3)) " for all values of " n in N`. |
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