find nth term of x to the power m
Answers
Answer:
Using the Binomial Theorem to Find a Single Term
Expanding a binomial with a high exponent such as \displaystyle {\left(x+2y\right)}^{16}(x+2y)
16
can be a lengthy process.
Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term.
Note the pattern of coefficients in the expansion of \displaystyle {\left(x+y\right)}^{5}(x+y)
5
.
\displaystyle {\left(x+y\right)}^{5}={x}^{5}+\left(\begin{array}{c}5\\ 1\end{array}\right){x}^{4}y+\left(\begin{array}{c}5\\ 2\end{array}\right){x}^{3}{y}^{2}+\left(\begin{array}{c}5\\ 3\end{array}\right){x}^{2}{y}^{3}+\left(\begin{array}{c}5\\ 4\end{array}\right)x{y}^{4}+{y}^{5}(x+y)
5
=x
5
+(
5
1
)x
4
y+(
5
2
)x
3
y
2
+(
5
3
)x
2
y
3
+(
5
4
)xy
4
+y
5
The second term is \displaystyle \left(\begin{array}{c}5\\ 1\end{array}\right){x}^{4}y(
5
1
)x
4
y. The third term is \displaystyle \left(\begin{array}{c}5\\ 2\end{array}\right){x}^{3}{y}^{2}(
5
2
)x
3
y
2
. We can generalize this result.
\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}(
n
r
)x
n−r
y
r
A GENERAL NOTE: THE (R+1)TH TERM OF A BINOMIAL EXPANSION
The \displaystyle \left(r+1\right)\text{th}(r+1)th term of the binomial expansion of \displaystyle {\left(x+y\right)}^{n}(x+y)
n
is:
\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}(
n
r
)x
n−r
y
r
HOW TO: GIVEN A BINOMIAL, WRITE A SPECIFIC TERM WITHOUT FULLY EXPANDING.
Determine the value of \displaystyle nn according to the exponent.
Determine \displaystyle \left(r+1\right)(r+1).
Determine \displaystyle rr.
Replace \displaystyle rr in the formula for the \displaystyle \left(r+1\right)\text{th}(r+1)th term of the binomial expansion.
EXAMPLE 3: WRITING A GIVEN TERM OF A BINOMIAL EXPANSION
Find the tenth term of \displaystyle {\left(x+2y\right)}^{16}(x+2y)
16
without fully expanding the binomial.
SOLUTION
Because we are looking for the tenth term, \displaystyle r+1=10r+1=10, we will use \displaystyle r=9r=9 in our calculations.
\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}(
n
r
)x
n−r
y
r
\displaystyle \left(\begin{array}{c}16\\ 9\end{array}\right){x}^{16 - 9}{\left(2y\right)}^{9}=5\text{,}857\text{,}280{x}^{7}{y}^{9}(
16
9
)x
16−9
(2y)
9
=5,857,280x
7
y
9
TRY IT 3
Find the sixth term of \displaystyle {\left(3x-y\right)}^{9}(3x−y)
9
without fully expanding the
Answer:
n^ √x^m
Step-by-step explanation:
if the base has power in fractions then denominator will turn into nth term of root