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The sequence f(n) = (2n)! ÷ n! is defined for all positive

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The sequence f(n) = (2n)! ÷ n! is defined for all positive  [#permalink]

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New post 15 Jan 2012, 15:28
6
25
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A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

66% (03:04) correct 34% (03:01) wrong based on 316 sessions

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The sequence f(n) = (2n)! ÷ n! is defined for all positive integer values of n. If x is defined as the product of the first 10 ten terms of this sequence, which of the following is the greatest factor of x?

(A) 2^20
(B) 2^30
(C) 2^45
(D) 2^52
(E) 2^55

The sequence f(n)=(2n)!/n! is defined for all positive integer values of n.
Guys - as the OA is not provided this is how I solved this. Can you please let me know if my approach is correct or not?

f(1) = 2
f(2) = \(2^2\) * 3
f(3) = \(2^3\) * 3 * 5
.
.
.
f(10) = 2^10

Product of 10 terms = 2^1 * 2^2 * ...........................2^10 --------------------------------(1)
The above can be simplified as the base is same i.e. 2

1+2+3+4+5+.........10= 2^55 and hence the answer is E

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The sequence f(n) = (2n)! ÷ n! is defined for all positive  [#permalink]

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New post 15 Jan 2012, 15:53
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enigma123 wrote:
The sequence f(n) = (2n)! ÷ n! is defined for all positive integer values of n. If x is defined as the product of the first 10 ten terms of this sequence, which of the following is the greatest factor of x?
(A) 2^20
(B) 2^30
(C) 2^45
(D) 2^52
(E) 2^55


Given: \(f(n)=\frac{(2n)!}{n!}\), so \(f(1)=\frac{(2)!}{1!}\), \(f(2)=\frac{(4)!}{2!}\), \(f(3)=\frac{(6)!}{3!}\), ...

\(x=\frac{(2)!}{1!}*\frac{(4)!}{2!}*\frac{(6)!}{3!}*...*\frac{(20)!}{10!}\).

Now, try to figure out the pattern in powers of 2 instead of brute force (reducing, factoring out 2's,...).

\(\frac{(2)!}{1!}=2^1\);

\(\frac{(4)!}{2!}=2^2*3\);

\(\frac{(6)!}{3!}=2^2*5*6=2^3*something\);

\(\frac{(8)!}{4!}=2^4*3*5*7=2^4*something\);

...

\(\frac{(20)!}{10!}=2^{10}*something\)

Basically the power of 2 goes up by 1 with each step.

So the greatest factor of x (greatest power of 2 which is a factor of x) will be \(2^1*2^2*2^1*...*2^{10}=2^{1+2+3+4+5+6+7+8+9+10}=2^{55}\) (mean*# of terms=5.5*10).

Answer: E.
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Re: The sequence f(n) = (2n)! ÷ n! is defined for all positive  [#permalink]

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New post 18 Jan 2019, 00:37
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Re: The sequence f(n) = (2n)! ÷ n! is defined for all positive   [#permalink] 18 Jan 2019, 00:37
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