P is the product, indicated above, of all the numbers of the form

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.

Let’s calculate some values from our set to find a pattern:

1 - ½ = ½

1 - ⅓ = ⅔

1 - ¼ = ¾

1 - ⅕ = ⅘

Multiplying these values together, we have:

(1/2)(2/3)(3/4)(4/5)

Notice that the fractions simplify and we are left with ⅕.

So, we are left with the numerator from the first fraction and the denominator from the last fraction.

For P, since the last fraction in the set is 1 - 1/50 = 49/50, P would equal 1/50.

Answer: B

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.

Another way to look at it is to turn P into a factorial.

(1) \(1 - \frac{1}{2} = \frac{1}{2}\)

(2) \(1 - \frac{1}{3} = \frac{2}{3}\)

(3) \(1 - \frac{1}{4} = \frac{3}{4}\)

(4) \(1 - \frac{1}{5} = \frac{4}{5}\)

\(P=\frac{1}{2} * \frac{2}{3} * \frac{3}{4} * \frac{4}{5} * . . . * \frac{49}{50}\)

∴ \(P=\frac{49!}{50!} = \frac{49!}{50*49!} = \frac{1}{50}\)

Answer B.

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(1-1/2)*(1-1/3)….(1-1/50)
1/2*2/3*…….49/50

1*2*3………*49
_____________ = 1/50 (B)
2*3*……….*50

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