Nathanlambson wrote:
Abhishek009 wrote:
. . .There is a pattern if you all can analyze -
\(P = (1-\frac{1}{2})\) = \(\frac{1}{2}\)
\(P = (1-\frac{1}{2})(1-\frac{1}{3})\) = \((\frac{1}{3})\)
\(P = (1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4})\) = \((\frac{1}{4})\)
Can you notice a pattern ?
[The pattern] depends on the value of the last denominator...
So, \(P = (1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4})...(1-\frac{1}{50})\) = \((\frac{1}{50})\)
Hence answer will be (B), \((\frac{1}{50})\)
So does this mean that if k was from 2 to 10 inclusive, the answer would be 1/10? I'm having a hard time following the math on this one.
Nathanlambson - yes, if k were 2 to 10 inclusive, the answer would be 1/10.
I had a hard time following the math, too. Perhaps how I rewrote the post might help.
1. From original equation, instead of considering all the factors in parentheses, take them
two at a time, and do any arithmetic that is inside parentheses.
\(P = (1-\frac{1}{2})(1-\frac{1}{3})\) =
\((\frac{1}{2})(\frac{2}{3})\) = \(\frac{2}{6}\) =
\(\frac{1}{3}\)2. Repeat, but put the end product from above as the first of your next two factors
\((\frac{1}{3})(1-\frac{1}{4})\) = \((\frac{1}{3})(\frac{3}{4})\) = \(\frac{1}{4}\)
And again, two at a time, use end product from previous step as first factor:
\((\frac{1}{4})(1-\frac{1}{5})\) = \((\frac{1}{4})(\frac{4}{5})\) = \(\frac{1}{5}\)
Abhishek009 wrote
"[The product of the preceding factors, at any stage, including the last calculation] depends on the value of the last denominator....
In the last step above, denominator is 5. 1 - (1/
5) = 4/
5 The pattern also demonstrates that after multiplication by previous term, the numerator is 1, and the product's denominator is 5. \((\frac{1}{4})(\frac{4}{5})\) = \(\frac{1}{5}\).
So the pattern is: the end product of all factors at any point including last term = 1 over whatever denominator the nth term has.
You could calculate k = 2 to 10 inclusive, easily, by multiplying
only the first two factors, and by inserting the answer to the arithmetic inside all the parentheses. Thus:
\((1-\frac{1}{2})(1-\frac{1}{3})
= (\frac{1}{2})(\frac{2}{3})\) = \(\frac{1}{3}\) ...
\(\frac{1}{3}\)*\(\frac{3}{4}\)*\(\frac{4}{5}\) *\(\frac{5}{6}\)*\(\frac{6}{7}\)* \(\frac{7}{8}\) *\(\frac{8}{9}\)*\(\frac{9}{10}\) =
Start canceling. You get \(\frac{1}{10}\)
Hope it helps.
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