A) \[\frac{{{a}_{2}}}{{{a}_{2}}+{{a}_{3}}}\]
B) \[\frac{1}{2}\frac{{{a}_{2}}}{({{a}_{2}}+{{a}_{3}})}\]
C) \[\frac{2{{a}_{2}}}{{{a}_{2}}+{{a}_{3}}}\]
D) \[\frac{2{{a}_{3}}}{{{a}_{2}}+{{a}_{3}}}\]
Correct Answer: C
Solution :
Let \[{{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}}\] be respectively the coefficients of \[{{(r+1)}^{th}},{{(r+2)}^{th}}\], \[{{(r+3)}^{th}}\] and \[{{(r+4)}^{th}}\]terms in the expansion of \[{{(1+x)}^{n}}\]. Then \[{{a}_{1}}={{\,}^{n}}{{C}_{r}},{{a}_{2}}={{\,}^{n}}{{C}_{r+1}},{{a}_{3}}={{\,}^{n}}{{C}_{r+2,}}{{a}_{4}}={{\,}^{n}}{{C}_{r+3}}\] Now \[\frac{{{a}_{1}}}{{{a}_{1}}+{{a}_{2}}}+\frac{{{a}_{3}}}{{{a}_{3}}+{{a}_{4}}}=\frac{^{n}{{C}_{r}}}{^{n}{{C}_{r}}+{{\,}^{n}}{{C}_{r+1}}}\]\[+\frac{^{n}{{C}_{r+2}}}{^{n}{{C}_{r+2}}+{{\,}^{n}}{{C}_{r+3}}}\] \[=\frac{^{n}{{C}_{r}}}{^{n+1}{{C}_{r+1}}}+\frac{^{n}{{C}_{r+2}}}{^{n+1}{{C}_{r+3}}}\]\[=\frac{^{n}{{C}_{r}}}{\frac{n+1}{r+1}{{\,}^{n}}{{C}_{r}}}+\frac{^{n}{{C}_{r+2}}}{\frac{n+1}{r+3}{{\,}^{n}}{{C}_{r+2}}}\]\[({{\because }^{n}}{{C}_{r}}=\frac{n}{r}{{\,}^{n-1}}{{C}_{r-1}})\] = \[\frac{r+1}{n+1}+\frac{r+3}{n+1}=\frac{2(r+2)}{n+1}\] \[=2\frac{^{n}{{C}_{r+1}}}{^{n+1}{{C}_{r+2}}}=2\frac{^{n}{{C}_{r+1}}}{^{n}{{C}_{r+1}}+{{\,}^{n}}{{C}_{r+2}}}=\frac{2{{a}_{2}}}{{{a}_{2}}+{{a}_{3}}}\]
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