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Using the principle of mathematical induction to show that `tan^(-1)(n+1)x-tan^(-1)x , forall x in N`. |
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Answer» Let `P(n):tan^(-1)((x)/(1+1.2.x^2))+tan^(-1)((x)/(1+2.3.x^2))+.....+tan^(-1)((x)/(1+n(n+1)x^2))` .....(i) `=tan^(-1)(n+1)x-tan^(-1)x` Step I For `n=1`. LHS of Eq. (i) `=tan^(-1)((x)/(1+1.2.x^2))` `=tan^(-1)((2x-x)/(1+2x.x))=tan^(-1)2x-tan^(-1)x` =RHS of Eq.(i) Therefore , P(1) is true. Step II Assume it is true for `n=k`. `P(k):tan^(-1)((x)/(1+1.2x^2))+tan^(-1)((x)/(1+2.3x^2))+......+tan^(-1)((x)/([1+k(k+1)x^2]))` `=tan^(-1)(k+1)x-tan^(-1)x` Step III For `n=k+1`. `P(k+1):tan^(-1)((x)/(1+1.2.x^2))+tan^(-1)((x)/(1+2.3.x^2))+.....+tan^(-1)((x)/(1+k(k+1)x^2))+......+tan^(-1)((x)/(1+(k=1)(k+2)x^2))` `=tan^(-1)(k+2)x-tan^(-1)x` `therefore LHS =tan^(-1)((x)/(1+1.2.x^2))` `+tan^(-1)((x)/(1+2.3.x^2))+....+tan^(-1)((x)/(1+k(k+1)x^2))+tan^(-1)((x)/(1+(k+1)(k+2)x^2))` `tan^(-1)(k+1)x-tan^(-1)x+tan^(-1)((x)/(1+(k+1)(k+2)x^2))` `=tan^(-1)(k+1)x-tan^(-1)x+tan^(-1)(((k+2)x-(k+1)x)/(1+(k+2)x(k+1)x))` `=tan^(-1)(k+1)x-tan^(-1)x+tan^(-1)(k+2)x-tan^(-1)(k+1)x=tan^(-1)(k+2)x-tan^(-1)x=RHS` This shows that the result is true for `n=k+1`. Hence , by the principle of mathematical induction , the result is true for all the `n in N`. |
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