A) 12
B) 10
C) 9
D) 8
Correct Answer: D
Solution :
| [d] We have, \[{{\left( 1+x+{{x}^{2}}+{{x}^{3}}+{{x}^{4}} \right)}^{n}}{{(x-1)}^{n+3}}\] |
| \[=\frac{{{({{x}^{5}}-1)}^{n}}}{{{(1-x)}^{n}}}.{{(x-1)}^{n+3}}={{({{x}^{5}}-1)}^{n}}{{(x-1)}^{3}}\] |
| \[=\left( +{{x}^{3}}-3{{x}^{2}}+3x-1 \right)\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{(-1)}^{r}}\,{{x}^{5r}}}\] |
| \[=+\sum\limits_{r=0}^{n}{{{\,}^{n}}{{C}_{r}}{{(-1)}^{r}}{{x}^{5r+3}}+3\sum\limits_{r=0}^{n}{{{\,}^{n}}{{C}_{r}}{{(-1)}^{r}}\,{{x}^{5r+2}}}}\] |
| \[-3\sum\limits_{r=0}^{n}{{{\,}^{n}}{{C}_{r}}{{(-1)}^{r}}\,{{x}^{5r+1}}+3\sum\limits_{r=0}^{n}{{{\,}^{n}}{{C}_{r}}}{{(-1)}^{r}}{{x}^{5r}}}\] |
| For term containing \[{{x}^{83}},\] we have \[5r+3=83\] |
| \[\Rightarrow r=16\] whereas \[5r+2=83,5r+1=83\] and \[5r=83\] give no integral value of r. hence, there is only one term containing \[{{x}^{83}}\] whose coefficient |
| \[=-{{\,}^{n}}{{C}_{16}}=-{{\,}^{n}}{{C}_{2\lambda }}\,\,\,\,\,\therefore \,\,2\lambda =16\Rightarrow \lambda =8\] |
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