Binomial Theorem And Its Simple Applications Question 94
Question: If $ P=n(n^{2}-1^{2})(n^{2}-2^{2})(n^{2}-3^{2})…(n^{2}-r^{2}), $ $ n>r,n\in N $ then P is necessarily divisible by
Options:
A) $ (2r+2)! $
B) $ (2r+4)! $
C) $ (2r+1)! $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] $ P=n(n+1)(n-1)(n+2)(n-2)…(n+r)(n-r). $
$ ={n(n+1)(n+2)…(n+r)}{(n-1)(n-2)…(n-r)} $
$ =(n+r)(n+r-1)…(n+1)(n)(n-1)…(n-r) $ Clearly P is product of $ (2r+1) $ consecutive integers, so divisible by $ (2r+1)! $
BETA
