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Question
Using the Mathematical induction, show that for any natural number n ≥ 2,
`(1 - 1/2^2)(1 - 1/3^2)(1 - 1/4^2) ... (1 - 1/"n"^2) = ("n" + 1)/2`
Solution
Let P(n) = is the statement `(1 -1/2^2)(1 -1/3^2) .... (1 - 1/"n"^2) = ("n" + 1)/2`
Given n ≥ 2
For n = 2
L.H.S = `(1 - 1/2^2)`
= `1 - 1/4`
= `3/4`
R.H.S P(2) = `(2 + 1)/(2(2))`
= `3/4`
L.H.S = R.H.S
P(n) is true for n = 2
Assume tat P(n) is true for n = k
(i.e.) `(1 -1/2^2)(1 -1/3^2) .... (1 - 1/"k"^2) = ("k" + 1)/2` is true
To prove P(k + 1) is true
Now P(k + 1) = `"P"("k") xx("t"_("k" + 1))`
= `("k" + 1)/(2"k") xx (1 - 1/("k" + 1)^2)`
= `("k" + 1)/(2"k") xx [(("k" + 1)^2 - 1)/("k" + 1)^2]`
= `("k" + 1)/(2"k") xx ("k"^2 + 2"k" + 1 - 1)/("k" + 1)^2`
= `("k" + 1)/(2"k") xx ("k"("k" + 2))/("k" + 1)^2`
= `("k" + 2)/(2("k" + 1))`
= `(("k" + 1 + 1))/(2("k" + 1))`
⇒ P(k + 1) is true when P(k) is true so by the principle of mathematical induction P(n) is true.
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