find the cofficient of x^7 in the expansion of( x^2 +1\x)^11.
Answers
Answer:
Coefficientofx
7
=462
Given \: (x^{2} + \frac{1}{x})^{11}Given(x
2
+
x
1
)
11
\begin{gathered}General \: term \:of \: binomial \: expansion \\(a+b)^{n} \: is\: t_{n}\end{gathered}
Generaltermofbinomialexpansion
(a+b)
n
ist
n
\boxed {\pink { t_{n} = ^nC_{r} \cdot a^{n-r}\cdot b^{r} }}
t
n
=
n
C
r
⋅a
n−r
⋅b
r
Here , a = x^{2} , b = \frac{1}{x} \:and \:n = 11Here,a=x
2
,b=
x
1
andn=11
t_{n} = ^{11} C_{r} \times (x^{2})^{11-r} \times (\frac{1}{x})^{r}t
n
=
11
C
r
×(x
2
)
11−r
×(
x
1
)
r
\begin{gathered}= ^{11}C_{r} \times x^{22-2r} \times x^{r}\\= ^{11}C_{r} \times x^{22-2r-r}\\= ^{11}C_{r}\times x^{22-3r} \:--(1)\end{gathered}
=
11
C
r
×x
22−2r
×x
r
=
11
C
r
×x
22−2r−r
=
11
C
r
×x
22−3r
−−(1)
/* According to the problem given,
22-3r = 7 \: (Given)22−3r=7(Given)
\implies -3r = 7 - 22⟹−3r=7−22
\implies r = \frac{-15}{-3} = 5⟹r=
−3
−15
=5
\begin{gathered}Coefficient \: of \: x^{7} = ^{11}C_{r} \\= ^{11}C_{5}\end{gathered}
Coefficientofx
7
=
11
C
r
=
11
C
5
\begin{gathered}= \frac{11!}{(11-5)! 5!}\\= \frac{11!}{6! 5! }\\= \frac{11\cdot10\cdot9\cdot8\cdot7\cdot6!}{6! \times 5\cdot4\cdot3\cdot2\cdot1 }\\= 462\end{gathered}
=
(11−5)!5!
11!
=
6!5!
11!
=
6!×5⋅4⋅3⋅2⋅1
11⋅10⋅9⋅8⋅7⋅6!
=462
Therefore.,
\red { Coefficient \:x^{7} }\green {= 462 }Coefficientx
7
=462