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30 May 2021, 23:20
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
76% (02:20) correct
24%
(02:52)
wrong
based on 95
sessions
History
Date
Time
Result
Not Attempted Yet
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) \(4n^2 − 20n + 24\)
A) 4
B) \(2n + 4\)
C) \(4n + 4\)
D) \(n^2 + 5n + 6\)
E) \(4n^2 − 20n + 24\)
30 May 2021, 23:27
Kudos
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sjuniv32
\(\frac{(n + 3)!}{(n + 1)!}\) − \(\frac{(n + 2)!}{n!}\) − \(\frac{(n + 1)!}{(n − 1)!}\) + \(\frac{n!}{(n − 2)! }\)
\(\frac{(n + 3)*(n+2)*(n+1)!}{(n + 1)!}\) − \(\frac{(n + 2)(n+1)n!}{n!}\) − \(\frac{(n + 1)n(n-1)!}{(n − 1)!}\) + \(\frac{n(n-1)(n-2)!}{(n − 2)! }\)
\((n+3)(n+2)-(n+1)(n+2)-(n+1)n+n(n-1)\)
\((n+2)(n+3-(n+1))-n(n+1-(n-1))\)
\((n+2)(n+3-n-1)-n(n+1-n+1)=(n+2)2-n(2)=2(n+2-n)=2*2=4\)
A
Substitute some value for n, say 15
\(\frac{(n + 3)!}{(n + 1)!}\) − \(\frac{(n + 2)!}{n!}\) − \(\frac{(n + 1)!}{(n − 1)!}\) + \(\frac{n!}{(n − 2)! }\)
\(\frac{18!}{16!}\) − \(\frac{17!}{15!}\) − \(\frac{16!}{14!}\) + \(\frac{15!}{13! }\)
\(18*17-17*16-16*15+15*14=17(18-16)+15(14-16)=17*2+15*-2=34-30=4\)
A
31 May 2021, 02:16
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chetan2u
Could you please explain the line below in an easier way? How did it come from the previous line? Thanks in advance!
\(\frac{(n + 3)*(n+2)*(n+1)!}{(n + 1)!}\) − \(\frac{(n + 2)(n+1)n!}{n!}\) − \(\frac{(n + 1)n(n-1)!}{(n − 1)!}\) + \(\frac{n(n-1)(n-2)!}{(n − 2)! }\)
Could you please explain the line below in an easier way? How did it come from the previous line? Thanks in advance!
\(\frac{(n + 3)*(n+2)*(n+1)!}{(n + 1)!}\) − \(\frac{(n + 2)(n+1)n!}{n!}\) − \(\frac{(n + 1)n(n-1)!}{(n − 1)!}\) + \(\frac{n(n-1)(n-2)!}{(n − 2)! }\)
