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If n is an integer, and n > 12, then (n + 3)!/(n + 1)! − (n + 2)!/n! −

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If n is an integer, and n > 12, then (n + 3)!/(n + 1)! − (n + 2)!/n! − [#permalink] New post   30 May 2021, 23:20
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If n is an integer, and n > 12, then \(\frac{(n + 3)!}{(n + 1)!}\) − \(\frac{(n + 2)!}{n!}\) − \(\frac{(n + 1)!}{(n − 1)!}\) + \(\frac{n!}{(n − 2)! }\) is equal to

A) 4

B) \(2n + 4\)

C) \(4n + 4\)

D) \(n^2 + 5n + 6\)

E) \(4n^2 − 20n + 24\)
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If n is an integer, and n > 12, then (n + 3)!/(n + 1)! − (n + 2)!/n! − [#permalink] New post   30 May 2021, 23:27
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sjuniv32 wrote:
If n is an integer, and n > 12, then \(\frac{(n + 3)!}{(n + 1)!}\) − \(\frac{(n + 2)!}{n!}\) − \(\frac{(n + 1)!}{(n − 1)!}\) + \(\frac{n!}{(n − 2)! }\) is equal to

A) 4

B) \(2n + 4\)

C) \(4n + 4\)

D) \(n^2 + 5n + 6\)

E) \(4n^2 − 20n + 24\)



\(\frac{(n + 3)!}{(n + 1)!}\) − \(\frac{(n + 2)!}{n!}\) − \(\frac{(n + 1)!}{(n − 1)!}\) + \(\frac{n!}{(n − 2)! }\)

\(\frac{(n + 3)*(n+2)*(n+1)!}{(n + 1)!}\) − \(\frac{(n + 2)(n+1)n!}{n!}\) − \(\frac{(n + 1)n(n-1)!}{(n − 1)!}\) + \(\frac{n(n-1)(n-2)!}{(n − 2)! }\)

\((n+3)(n+2)-(n+1)(n+2)-(n+1)n+n(n-1)\)

\((n+2)(n+3-(n+1))-n(n+1-(n-1))\)

\((n+2)(n+3-n-1)-n(n+1-n+1)=(n+2)2-n(2)=2(n+2-n)=2*2=4\)

A

Substitute some value for n, say 15

\(\frac{(n + 3)!}{(n + 1)!}\) − \(\frac{(n + 2)!}{n!}\) − \(\frac{(n + 1)!}{(n − 1)!}\) + \(\frac{n!}{(n − 2)! }\)

\(\frac{18!}{16!}\) − \(\frac{17!}{15!}\) − \(\frac{16!}{14!}\) + \(\frac{15!}{13! }\)

\(18*17-17*16-16*15+15*14=17(18-16)+15(14-16)=17*2+15*-2=34-30=4\)

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If n is an integer, and n > 12, then (n + 3)!/(n + 1)! − (n + 2)!/n! − [#permalink] New post   31 May 2021, 02:16
chetan2u
Could you please explain the line below in an easier way? How did it come from the previous line? Thanks in advance!

\(\frac{(n + 3)*(n+2)*(n+1)!}{(n + 1)!}\) − \(\frac{(n + 2)(n+1)n!}{n!}\) − \(\frac{(n + 1)n(n-1)!}{(n − 1)!}\) + \(\frac{n(n-1)(n-2)!}{(n − 2)! }\)
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Re: If n is an integer, and n > 12, then (n + 3)!/(n + 1)! − (n + 2)!/n! − [#permalink] New post   31 May 2021, 09:06
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sjuniv32 wrote:
chetan2u
Could you please explain the line below in an easier way? How did it come from the previous line? Thanks in advance!

\(\frac{(n + 3)*(n+2)*(n+1)!}{(n + 1)!}\) − \(\frac{(n + 2)(n+1)n!}{n!}\) − \(\frac{(n + 1)n(n-1)!}{(n − 1)!}\) + \(\frac{n(n-1)(n-2)!}{(n − 2)! }\)



Hi
n! or factorial means product of all positive integers till n.

(n+3)! will mean (n+3)(n+2)(n+1)(n)(n-1)......4*3*2*1
This can be further written as (n+3)!=(n+3)*(n+2)!=(n+3)(n+2)*(n+1)! and so on.

I have expanded the numerator in such a way that the denominator can be canceled out.

Hope I have understood your query well, and it clears the query.
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If n is an integer, and n > 12, then (n + 3)!/(n + 1)! − (n + 2)!/n! − [#permalink] New post   31 May 2021, 09:41
chetan2u , Thanks for the above explanation. I got the explanation of the first fraction bar and I experienced difficulty in the other three fraction bars. I think I have got it now, but it would confirm my understanding if you explained the one below:

How \(\frac{n!}{(n−2)!}\) becomes \(\frac{n(n-1)(n-2)!}{(n − 2)! }\) ?
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Re: If n is an integer, and n > 12, then (n + 3)!/(n + 1)! − (n + 2)!/n! − [#permalink] New post   31 May 2021, 09:54
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sjuniv32 wrote:
chetan2u , Thanks for the above explanation. I got the explanation of the first fraction bar and I experienced difficulty in the other three fraction bars. I think I have got it now, but it would confirm my understanding if you explained the one below:

How \(\frac{n!}{(n−2)!}\) becomes \(\frac{n(n-1)(n-2)!}{(n − 2)! }\) ?


n! =1*2*3......*(n-2)*(n-1)*n= n*(n-1)*(n-2)*.....3*2*1=n(n-1)(n-2)!

So n!/(n-2)!=n(n-1)(n-2)!/(n-2)!=n(n-1)
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Re: If n is an integer, and n > 12, then (n + 3)!/(n + 1)! − (n + 2)!/n! − [#permalink]
31 May 2021, 09:54
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