y7214001 wrote:
Kudos +1 Karishma
Is there an fast way to compute the result of the multiplacation series like we have for \(
Cf\)? I actually did the long way .
It would have taken forever!
\(C_f = 1 * (\frac{2}{4}) * (\frac{3}{5}) * (\frac{4}{6}) * (\frac{5}{7}) * ....... (\frac{18}{20}) * (\frac{19}{21}) * (\frac{20}{22})\)
You need to observe here that other than first two numerators and last two denominators, all other terms will cancel out.
First term's denominator will cancel out third term's numerator.
Second term's denominator will cancel out fourth term's numerator.
The last two denominators will have no numerators to cancel them out.
The first two numerators have no denominators to cancel them out.
Usually, in such expressions (where terms have a pattern), things simplify easily. You just need to observe the pattern.
_________________
Karishma
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