A) n divides 3S
B) n+1 divides 3S
C) n+2 divides 3S
D) All are correct
Correct Answer: D
Solution :
| [d] We have, \[S=\sum\limits_{k=0}^{n-1}{^{k+2}{{P}_{2}}=2!\sum\limits_{k=0}^{n-1}{^{k+2}{{C}_{2}}}}\] |
| \[=(2){{[}^{2}}{{C}_{2}}{{+}^{3}}{{C}_{2}}{{+}^{4}}{{C}_{2}}+...{{+}^{n+1}}{{C}_{2}}]\] |
| 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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