Problems Set at an Oral Examination.
Prove that the following equations hold for any natural n.
$(1 - \frac{1}{4}) (1 - \frac{1}{9}) (1 - \frac{1}{16}) . . . (1 - \frac{1}{n^{2}}) = \frac{n + 1}{2 n} .$

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Solution

$1 - \frac{1}{n^{2}} = \frac{n^{2} - 1}{n^{2}} = \frac{(n - 1) (n + 1)}{n^{2}} = \frac{n - 1}{n} . \frac{n + 1}{n}$

$1 - \frac{1}{4} = \frac{1}{2} . \frac{3}{2}$

$1 - \frac{1}{9} = \frac{2}{3} . \frac{4}{3}$

$1 - \frac{1}{16} = \frac{3}{4} . \frac{5}{4}$

$1 - \frac{1}{25} = \frac{4}{5} . \frac{6}{5}$

And so on...

Hence, we get-

$\frac{1}{2} . \frac{3}{2} . \frac{2}{3} . \frac{4}{3} . \frac{3}{4} . \frac{5}{4} . \frac{4}{5} . \frac{6}{5} . . . \frac{n - 1}{n} . \frac{n + 1}{n} .$

$= \frac{1}{2} . (\frac{3}{2} . \frac{2}{3}) (\frac{4}{3} . \frac{3}{4}) . (\frac{5}{4} . \frac{4}{5}) . (\frac{6}{5} . . . \frac{n - 1}{n}) . \frac{n + 1}{n} .$

$= \frac{1}{2} .1 .1 .1 . . . . \frac{n + 1}{n} = \frac{1}{2} \frac{n + 1}{n}$

$= \frac{n + 1}{2 n}$

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Problems Set at an Oral Examination.Prove that the following equations hold for any natural n.(1−14)(1−19)(1−116)...(1−1n2)=n+12n.

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Problems Set at an Oral Examination.
Prove that the following equations hold for any natural n.
$(1 - \frac{1}{4}) (1 - \frac{1}{9}) (1 - \frac{1}{16}) . . . (1 - \frac{1}{n^{2}}) = \frac{n + 1}{2 n} .$