A) \[1\]
B) \[2\]
C) \[3\]
D) \[4\]
Correct Answer: C
Solution :
[c] \[{{S}_{n}}{{=}^{n}}{{C}_{0}}{{.}^{n}}{{C}_{n}}{{+}^{n}}{{C}_{1}}{{.}^{n}}{{C}_{n-1}}{{+}^{n}}{{C}_{2}}{{.}^{n}}{{C}_{n-2}}+....\] \[{{+}^{n}}{{C}_{n}}{{.}^{n}}{{C}_{0}}\] \[{{=}^{n}}{{C}_{0}}{{.}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}{{.}^{n}}{{C}_{1}}{{+}^{n}}{{C}_{2}}{{.}^{n}}{{C}_{2}}+.....{{+}^{n}}{{C}_{n}}{{.}^{n}}{{C}_{n}}\] \[={{2}^{n}}{{C}_{n}}\] \[\Rightarrow \,\,\,\frac{{{S}_{n+1}}}{{{S}_{n}}}=\frac{^{2n+2}{{C}_{n+1}}}{^{2n}{{C}_{n}}}=\frac{(2n+2)\,(2n+1)}{(n+1)(n+1)}\] \[=\frac{2(2n+1)}{n+1}=4-\frac{2}{n+1}\] \[\therefore \,\,\,\,{{\left[ \frac{{{S}_{n+1}}}{{{S}_{n}}} \right]}_{\max }}=3\]You need to login to perform this action.
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