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Using mathematical induction , show that...

Using mathematical induction , show that `(1-(1)/(2^2))(1-(2)/(3^2))(1-(1)/(4^2)).....(1-(1)/((n+1)^2))=(n+2)/(2(n+1)), forall n in N`.

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Let `P(n):(1-(1)/(2^2))(1-(1)/(3^2))(1-(1)/(4^2))....(1-(1)/((n+1)^2))=(n+2)/(2(n+1))` ......(i)
Step I For `n=1`,
LHS of Eq. (i) `=1-(1)/(2^2)=(3)/(4)` and RHS of Eq. (i) `=(3)/(2.2)=(3)/(4)`
Therefore , P(1) is true.
Step III For `n=k+1`.
`P(k+1):(1-(1)/(2^2))(1-(1)/(3^2))(1-(1)/(4^2))(1-(1)/((k+1)^2))(1-(1)/((k+2)^2))=(k+3)/(2(k+2))`
`therefore LHS=(1-(1)/(2^2))(1-(1)/(3^2))(1-(1)/(4^2))....(1-(1)/((k+1)^2))(1-(1)/((k+2)^2))`
`=(k+2)/(2(k+1))(1-(1)/((k+2)^2))` [by assumption step]
`=((k+2))/(2(k+1)).([(k+2)^2-1])/((k+2)^2)=(k^2+4k+3)/(2(k+1)(k+2))`
`=((k+1)(k+3))/(2(k+1)(k+2))=((k+3))/(2(k+2))=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 `n in N`.
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