InterviewSolution
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InterviewSolution
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For `a gt 01`, prove thjat `(1+a)^(n) ge (1+an)` for all natural numbers n. |
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Answer» When n=1 LHS=(1+a) and RHS =(1+a) Hence `(1+a)^(n) ge (1+na)` is true, when n=1 `therefore P(1)` is true. (a) Let P(k) be true `Rightarrow (1+a)^(k) ge (1+ka)`…..(i) To prove P(k+1) is true, i.e. `(1+a)^(k+1) ge 1+(k+1)a=x("say")` ......(ii) `"from "(i), (1+a)^(k+1) ge (1+ka)(1+a)y(say)` Now, (y-x)=`(1+ka) +a+ka^(2)-1-ka-a)` `=1+ka+a+ka^(2)-1-1ka-a` `=ka^(2) gt 0 therefore y gt x` From (iii) and (iv), we get `(1+a)^(k+1) ge 1,+(k+1)a` Hence, P(k+1) is true whenever P(k) is true......(b) From (a) and (b) by the principle of mathematical induction, it follows that P(n) is true for all natural number n. |
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