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`1^2/(1.3)+2^2/(3.5)+3^2/(5.7)+.....+n^2/((2n-1)(2n+1))=((n)(n+1))/((2(2n+1))` |
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Answer» Let `P(n):(1^2)/(1:3)+(2^2)/(3.5)+.....+(n^2)/((2n-1)(2n+1))=(n(n+1))/(2(2n+1))` .....(i) Step I For n=1. LHS of Eq. (i) `(1^2)/(1.3)=(1)/(3)` RHS of Eq. (i) `=(1(1+1))/(2(2xx1+1))=(2)/(2(3))=(1)/(3)` LHS=RHS Therefore , P(1) is true. Step II Let us assume that the result is true for `n=k`, then `P(k)=(1^2)/(1.3)+(2^2)/(3.5)+....+(k^2)/((2k-1)(2k+1))=(k(k+1))/(2(2k+1))` Step III For `n=k+1`, we have to prove that `P(k+1):(1^2)/(1:3)+(2^2)/(3.5)+......+(k^2)/((2k-1)(2k+1))+((k+1)^2)/((2k+1)(2k+3))=((k+1)(k+2))/(2(2k+3))` LHS `=(1^2)/(1.3)+(2^2)/(3.5)+......+(k^2)/((2k-1)(2k+1))+((2k+1)^2)/((2k+1)(2k+1))` `=(k(k+1))/(2(2k+1))+((k+1)^2)/((2k+1)(2k+3))` [ by assumption step] `=((k+1))/((2k+1)){(k)/(2)+(k+1)/((2+3))}=((k+1))/((2k+1)){(2k^2+5k+2)/(2(2k+3))}` `=((k+1))/((2k+1)).((k+2)(2k+1))/(2(2k+3))=((k+1)(k+2))/(2(2k+3))=RHS` This shows that ,the result is true for `n=k+1`. Therefore , by the principle of mathematical induction the result is true for all ` n in N`. |
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