A) \[\frac{n(n+1)....(n-r+1)}{r!}{{a}^{n-r+1}}{{(2x)}^{r}}\]
B) \[\frac{n(n-1)....(n-r+2)}{(r-1)\,!}{{a}^{n-r+1}}{{(2x)}^{r-1}}\]
C) \[\frac{n(n+1)....(n-r)}{(r+1)!}{{a}^{n-r}}{{(x)}^{r}}\]
D) None of these
Correct Answer: B
Solution :
\[{{r}^{th}}\] term of \[{{(a+2x)}^{n}}\]is \[^{n}{{C}_{r-1}}{{(a)}^{n-r+1}}{{(2x)}^{r-1}}\] \[=\frac{n!}{(n-r+1)!(r-1)!}{{a}^{n-r+1}}{{(2x)}^{r-1}}\] \[=\frac{n(n-1).....(n-r+2)}{(r-1)!}{{a}^{n-r+1}}{{(2x)}^{r-1}}\]
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