The 12th term in the expansion of `e^(-3)` is

Options

A. `3^(11)/(11!)`
B. `(-3^(11))/(11!)`
C. `(11^(11))/(3!)`
D. `(-11^(11))/(3^(11))`

Answer

To find the 12th term in the expansion of \( e^{-3} \), we can use the Taylor series expansion for \( e^x \):

\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]

In our case, we have \( x = -3 \). Therefore, the expansion becomes:

\[ e^{-3} = \sum_{n=0}^{\infty} \frac{(-3)^n}{n!} \]

The \( n \)-th term of the series is given by:

\[ T_n = \frac{(-3)^n}{n!} \]

To find the 12th term, we need to substitute \( n = 11 \) (since the first term corresponds to \( n = 0 \)):

\[ T_{12} = \frac{(-3)^{11}}{11!} \]

Now, we calculate \( (-3)^{11} \) and \( 11! \):

1. Calculate \( (-3)^{11} \):
\[ (-3)^{11} = -3^{11} \]

2. Calculate \( 11! \):
\[ 11! = 39916800 \]

Now we can substitute these values back into the term:

\[ T_{12} = \frac{-3^{11}}{11!} \]

Finally, we can express the answer as:

\[ T_{12} = \frac{-3^{11}}{39916800} \]

Thus, the 12th term in the expansion of \( e^{-3} \) is:

\[ \frac{-3^{11}}{39916800} \]