Q.
Let n=2015 The least positive integer k for which kn2n2−12n2−22n3−32…n2−(n−1)2=r! for some positive integer r is
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a
2014
b
2013
c
1
d
2
answer is D.
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Detailed Solution
We can rewrite the given expression as kn2(n−1)(n+1)(n−2)(n+2)(n−3)(n+3)……….(n+n−1)(n−n+1)=r!⇒kn(1)(2)…(n−1)n(n+1)(n+2)…(2n−1)=r!⇒kn(2n−1)!=r!∴ To convert L.H.S. to a factorial, we shall requirek = 2 which will convert it into (2n)!
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