The coefficient of `x^(10)` in the series of `e^(2x)` is

Options

A. `(2^(9))/(9!)`
B. `(2^(10))/(10!)`
C. `(1)/(10!)`
D. none of these

Explanation

We have
`e^(x)=1+(x)/(1!)+(x^(2))/(2!)+(x^(3))/(3!)+…= underset(n=0)overset(infty)Sigma(2x)^(n)/(n!)`
Replacing x by 2x we get
`e^(2x)=1+(2x)/(1!)+(2x)^(2)/(2!)+(2x)^(3)/(3!)+….underset(n=0)overset(infty)Sigma(2x)^(n)/(n!)`
Clearly
`T_(n+1)=(n+1)^(th) term =(2x)^(n)/(n!)=(2^(n))/(n!)x^(n)`
If `(n+1)^(th)` term contains `x^(10)` then we must have n =10 ltbvrgt `therefore` Coefficient of `x^(10)=(2^(10))/(10!)`