A) \[^{n+m+2}{{C}_{n}}\]
B) \[^{n+m+3}{{C}_{n-1}}\]
C) \[{{x}^{-7}}\]
D) \[{{\left[ ax-\frac{1}{b{{x}^{2}}} \right]}^{11}}\]
Correct Answer: B
Solution :
Given \[\sin n\theta =\sum\limits_{r=0}^{n}{{{b}_{r}}{{\sin }^{r}}\theta }\] \[\Rightarrow \]\[\sin n\theta ={{b}_{0}}{{\sin }^{0}}\theta +{{b}_{1}}{{\sin }^{1}}\theta +{{b}^{2}}{{\sin }^{2}}\theta \] \[+....+{{b}_{n}}{{\sin }^{n}}\theta \] \[\Rightarrow \]\[\sin n\theta ={{b}_{0}}+{{b}_{1}}\sin \theta +....+{{b}^{n}}{{\sin }^{n}}\theta \] \[\because \]\[\sin n\theta {{=}^{n}}{{C}_{1}}\sin \theta {{\cos }^{n-1}}\theta \] \[{{-}^{n}}{{C}_{3}}{{\sin }^{3}}\theta {{\cos }^{n-3}}\theta +......\] \[{{=}^{n}}{{C}_{1}}\sin \theta {{(1-{{\sin }^{2}}\theta )}^{(n-1)/2}}\] \[{{-}^{n}}{{C}_{3}}{{\sin }^{3}}\theta {{(1-{{\sin }^{2}}\theta )}^{(n-3)/2}}+....\] \[\therefore \] \[{{b}_{0}}=0,\] \[{{b}_{1}}=\]coefficient of\[\sin \theta {{=}^{n}}{{C}_{1}}=n\] (\[\because \]\[n-1,\text{ }n-3\]are all even integer)
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