InterviewSolution
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InterviewSolution
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Show by using the principle of mathematical induction that for all natural number `n gt 2, 2^(n) gt 2n+1` |
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Answer» Let `P(n): 2^(n) gt 2n+1`, where `n gt 2` . When n=3, LHS=`2^(3)=8 and RHS=2xx3+1=7` clearly `8 gt 7` therefore `P(3)` is true. (a) Let P(k) be true In (i), multiplying both sides by 2, we get `2^(k+1) gt 4k+2…… (iii)` Now, (4k+2)-(2k+3)=`2k-1 gt 0` `Rightarrow 4k+2 gt 2k+3 ....(iv)` From (iii) and (iv), we get `2^(k+1) gt 2k+3` Hence P(k+1) is true whenever P(k) is true, From (a) and (b), it follows that P(n) is true for all natual number n. |
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