Question

Find the coefficient of x^7 in the expansion of (x^2+1/x)^11

Answer 1

Answer:

$\red { Coefficient \:of \:x^{7} }\green {= 462}$

Step-by-step explanation:

$Given \: \left(x^{2} + \frac{1}{x}\right)^{11}$

$General \: term \:of \: binomial \: expansion \\(a+b)^{n} \: is\: t_{n}$

$\boxed {\pink { t_{n} = ^nC_{r} \cdot a^{n-r}\cdot b^{r} }}$

$Here , a = x^{2} , b = \frac{1}{x} \:and \:n = 11$

$t_{n} = ^{11} C_{r} \times (x^{2})^{11-r} \times \left(\frac{1}{x}\right)^{r}$

$= ^{11}C_{r} \times x^{22-2r} \times x^{r}\\= ^{11}C_{r} \times x^{22-2r-r}\\= ^{11}C_{r}\times x^{22-3r} \:--(1)$

/* According to the problem given,

$22-3r = 7 \: (Given)$

$\implies -3r = 7 - 22$

$\implies r = \frac{-15}{-3} = 5$

$Coefficient \: of \: x^{7} = ^{11}C_{r} \\= ^{11}C_{5}$

$= \frac{11!}{(11-5)! 5!}\\= \frac{11!}{6! 5! }\\= \frac{11\cdot10\cdot9\cdot8\cdot7\cdot6!}{6! \times 5\cdot4\cdot3\cdot2\cdot1 }\\= 462$

Therefore.,

$\red { Coefficient \:x^{7} }\green {= 462 }$

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Answer 2

Answer:

$\displaystyle \red{{\rightarrow 462}}$

Step-by-step explanation:

Given :

$\displaystyle{\left(x^{7}-\frac{1}{x}\right)^{11}}\\\\\\\display \text{We have : }\\\\\\\displaystyle{\text{T}}_{r+1}= \ ^{n}\text{C}_r \ a^{n-r} \ b^{r}}\\\\\\\displaystyle{\text{T}}_{6+1}= \ ^{11}\text{C}_6 \ x^{11-6} \ 1/x^{6}}\\\\\\\displaystyle{\text{T}}_{7}= \ ^{11}\text{C}_6 \ 1/ x}\\\\\\\display \text{Now coefficient of x^{7} }\\\\\\\displaystyle{\implies \frac{11!}{(11-6)!\times 6!} }\\\\\\\displaystyle{\implies \frac{11!}{5!\times 6!} }$

$\displaystyle{\implies \frac{11\times10\times9\times8\times7\times6!}{5\times4\times3\times2\times1\times 6!} }\\\\\\\displaystyle{\implies 11\times2\times3\times7}\\\\\\\displaystyle{ \implies 22\times21}\\\\\\\displaystyle{ \implies 462}\\\\\\\display \text{Therefore , the coefficient of x^7 is 462 in expansion of \left(x+\frac{1}{x} \right)^{11}. }$