Permutations And Combinations Question 373
Question: Let $ S=\sum\limits _{k=0}^{n-1}{^{k+2}P_2,} $ then
Options:
A) n divides 3S
B) n+1 divides 3S
C) n+2 divides 3S
D) All are correct
Correct Answer: D $ =(2){{[}^{2}}C_2{{+}^{3}}C_2{{+}^{4}}C_2+…{{+}^{n+1}}C_2] $
But $ ^{2}C_2{{+}^{3}}C_2{{+}^{4}}C_2+…{{+}^{n+1}}C_2 $ $ ={{(}^{3}}C_3{{+}^{3}}C_2){{+}^{4}}C_2+…{{+}^{n+1}}C_2 $ $ ={{(}^{4}}C_3{{+}^{4}}C_2){{+}^{5}}C_2+…{{+}^{n+1}}C_2 $ $ ={{}^{5}}C_3+{{}^{5}}C_2+…{{+}^{n+1}}C_2 $ $ {{=}^{n+2}}C_3=\frac{1}{6}n(n+1)(n+2) $ $ \Rightarrow 3S=n(n+1)(n+2) $ $ \Rightarrow $ n divides 3S, (n+1) divides 3S and (n+2) divides 3S.
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Answer:
Solution:
BETA
