A) 1
B) 5
C) 25
D) 100
Correct Answer: A
Solution :
The numerator is of the form \[{{a}^{3}}+{{b}^{3}}+3ab(a+b)={{(a+b)}^{3}}\] \[\therefore {{N}^{r}}={{(18+7)}^{3}}={{25}^{3}}\] \[\therefore {{D}^{r}}={{3}^{6}}+{{\,}^{6}}{{C}_{1}}{{3}^{5}}{{.2}^{1}}+{{\,}^{6}}{{C}_{2}}{{3}^{4}}{{.2}^{2}}\]\[{{3}^{2n+2}}-8n-9,\,\forall n\in N\] This is clearly the expansion of \[{{(3+2)}^{6}}={{5}^{6}}={{(25)}^{3}}\] \[\therefore \,\,\,\frac{{{N}^{r}}}{{{D}^{r}}}=\frac{{{(25)}^{3}}}{{{(25)}^{3}}}=1\]
You need to login to perform this action.
You will be redirected in
3 sec
