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Question
Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:
`(1 - 1/2^2).(1 - 1/3^2)...(1 - 1/n^2) = (n + 1)/(2n)`, for all natural numbers, n ≥ 2.
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Solution
Let the given statement be P(n)
i.e., P(n) : `(1 - 1/2^2).(1 - 1/3^2)...(1 - 1/n^2) = (n + 1)/(2n)`, for all natural numbers, n ≥ 2
We, observe that P(2) is true
Since `(1 - 1/2^2) = 1 - 1/4`
= `(4 - 1)/4`
= `3/4`
= `(2 + 1)/(2 xx 2)`
Assume that P(n) is true for some k ∈ N
i.e., P(k) : `1 - 1/2^2 . 1 - 1/3^2 ... 1 - 1/k^2 = (k + 1)/(2k)`
Now, to prove that P(k + 1) is true,
We have `1 - 1/2^2 * 1 - 1/3^2 ... 1 - 1/k^2 . 1 - 1/(k + 1)^2`
= `(k + 1)/(2k)(1 - 1/(k + 1)^2)`
= `(k^2 + 2k)/(2k(k + 1))`
= `((k + 1) + 1)/(2(k + 1))`
Thus, P(k + 1) is true, Whenever P(k) is true.
Hence, by the Principle of Mathematical Induction, P(n) is true for all natural numbers, n ≥ 2.
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