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PROVE THAT (3n)! is divisible by n!*(n+1)!*(n+2)!.
Hi Aritra,
This is not true.
Check that for n=1, n=2....... (3n)! will not be divisible by n!(n+1)!(n+2)!....
This result holds true only for n ≥ 3, and is quite easy to prove.
Use that the product of n-consecutive numbers, is always divisible by n!.
Hope this helps.
Best Regards,
Ashwin (IIT Madras).
Dear aritra,
(3n)! = (3n)(3n-1)(3n-2)(3n-3)(3n-4)(3n-5)(3n-6)...............(2n)(2n-1)(2n-2)(2n-2)(2n-3).......(n+2)(n+1)n!----1
take the terms 3n,3n-3,3n-6,3n-9.....upto 3 and 2n,2n-2,2n-4.........upto 2
together and take 3 common from each and similarly 2 common from each
so we get n! twice and there is one more n! is already present so when we divide n!*(n+1)!*(n+2)!
all the three n! gets cancelled and remaining n+1, n+2 is also present in the equation 1
so n!*(n+1)!*(n+2)! divides 3n! completely ------hence proved
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jitender
askiitians expert
The problem is wrong.Take an example and verify.It us nt verified.
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