A) \[8\]
B) \[7\]
C) \[4\]
D) \[6\]
Correct Answer: D
Solution :
Given, \[{{(1+x)}^{15}}\] Now, \[{{T}_{r+1}}{{=}^{15}}{{C}_{r}}{{x}^{r}}\] and \[{{T}_{(r+3)+1}}{{=}^{15}}{{C}_{r+3}}{{x}^{r+3}}\] According to questions Coefficient of \[{{x}^{r}}\] = coefficient of \[{{x}^{r+3}}\] \[\Rightarrow \] \[^{15}{{C}_{r}}{{=}^{15}}{{C}_{r+3}}\] \[\Rightarrow \] \[\frac{15!}{r!(15-r)!}=\frac{15!}{(r+3)!(12-r)!}\] \[\Rightarrow \] \[\frac{1}{(15-r)(14-r)(13-r)}\] \[=\frac{1}{(r+3)(r+2)(r+1)}\] \[\Rightarrow \]\[(r+1)(r+2)(r+3)=(15-r)\] \[(14-r)(13-r)\] \[\Rightarrow \] \[({{r}^{2}}+3r+2)(r+3)=(210-29r+{{r}^{2}})\] \[=(13-r)\] \[\Rightarrow \] \[{{r}^{3}}+3{{r}^{2}}+2r+3{{r}^{2}}+9r+6\] \[=2930-377r+13{{r}^{2}}-210r+29{{r}^{2}}-{{r}^{3}}\] \[\Rightarrow \] \[2{{r}^{3}}-36{{r}^{2}}+598r-2924=0\] \[\Rightarrow \] \[{{r}^{3}}-18{{r}^{2}}+299r-1462=0\] \[\Rightarrow \] \[(r-6)({{r}^{2}}-12r+227)=0\] \[\Rightarrow \] \[r=6\] and \[{{r}^{2}}-12r+227=0\] gives imaginary roots. Alternate Method \[^{15}{{C}_{r}}{{=}^{15}}{{C}_{r+3}}\] \[\Rightarrow \] \[r+(r+3)=15\] (\[(\because \,{{\,}^{n}}{{C}_{x}}{{=}^{n}}{{C}_{y}})\] \[\Rightarrow \] \[2r+3=15\] \[\Rightarrow \] \[x+y=n\] \[\Rightarrow \] \[2r=12\] \[\Rightarrow \] \[r=6\]
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