Question

Find the coefficient x^7 in
\[\huge (ax^{2}+\frac{ 1 }{ bx })^{11}\]
and of x^ -7 in
\[\huge (ax-\frac{ 1 }{ bx^{2} })^{11}\]
and find the relation between "a" and "b" so that these coefficients are equal (where \[a,b \neq 0\]

Answer 1

OK - I don't know what you did, and I might have taken the longer route, but what is the answer given?

Answer 2

ab=1
or a=1/b

Answer 3

in the expansion of \[\left( x+a \right)^n\]
General term is
\[T _{r+1} =C_{r}^{n}x ^{n-r}a ^{r}\]

Answer 4

yes

Answer 5

\[here x=ax^2,a= \frac{ 1 }{ bx }\]

Answer 6

yes agreed

Answer 7

\[T _{r+1}=C _{r}^{11}\left( a x^2 \right)^{11-r}\left( \frac{ 1 }{ bx } \right)^r\]
\[=C _{r}^{11}a ^{11-r}x ^{22-2r}\frac{ 1 }{ b^rx^r }=C _{r}^{11}a ^{11-r}b ^{-r}x ^{22-2r-r}\]
\[=C _{r}^{11}a ^{11-r}b ^{-r}x ^{22-3r}\]
put 22-3r=7
3r=22-7=15
r=5
now find the coefficient.

Answer 8

i made a mistake i didn't take x^2

Answer 9

It is \[C _{5}^{11}a ^{11-5}b ^{-5}=?\]

Answer 10

\[\huge C_{5}^{11}\]

Answer 11

that's the coefficient

Answer 12

no, a and b are also constants,any thing other than x and its power is coefficient.

Answer 13

\[\huge C _{5}^{11}a^{6}b^{-5}\]

Answer 14

well negative 1/bx isn't there

Answer 15

its postive

Answer 16

now i have seen it is in second question.

Answer 17

in second \[a=-\frac{ 1 }{ bx^2 }\]

Answer 18

we are doing first right

Answer 19

correct

Answer 20

okay i understand the situation , well , then i get
\[\huge C _{5}^{11}a^{6}b^{-5}\]

Answer 21

correct.

Answer 22

\[C _{5}^{11}=\frac{ 11*10*9*8*7 }{ 5*4*3*2*1 }=?\]

Answer 23

132

Answer 24

now solve second as we solved first then equate coefficients.

Answer 25

doing it

Answer 26

I think 132 is not correct.

Answer 27

\[\left(a x^2+\frac{1}{b x}\right)^{11}=a^{11}
x^{22}+\frac{11 a^{10} x^{19}}{b}+\frac{55
a^9 x^{16}}{b^2}+\frac{165 a^8
x^{13}}{b^3}+\frac{330 a^7
x^{10}}{b^4}+\\\frac{462 a^6
x^7}{b^5}+\frac{462 a^5 x^4}{b^6}+\frac{330
a^4 x}{b^7}+\frac{165 a^3}{b^8 x^2}+\frac{55
a^2}{b^9 x^5}+\frac{11 a}{b^{10}
x^8}+\frac{1}{b^{11} x^{11}}\]

Answer 28

it is 462
so coefficient is \[462~a^6~b ^{-5}\]

Answer 29

for second x=ax

Answer 30

\[\left(a x-\frac{1}{b x^2}\right)^{11}=a^{11}
x^{11}-\frac{11 a^{10} x^8}{b}+\frac{55 a^9
x^5}{b^2}-\frac{165 a^8 x^2}{b^3}+\frac{330
a^7}{b^4 x}-\frac{462 a^6}{b^5 x^4}+\\\frac{462
a^5}{b^6 x^7}-\frac{330 a^4}{b^7
x^{10}}+\frac{165 a^3}{b^8 x^{13}}-\frac{55
a^2}{b^9 x^{16}}+\frac{11 a}{b^{10}
x^{19}}-\frac{1}{b^{11} x^{22}}

\]

Answer 31

\[
\frac{462 a^5}{b^6}=\frac{462 a^6}{b^5}
\]
you get a b=1

Answer 32

that is a long procedure ,better take a general term as we did in first.

Answer 33

Sorry I miss the minus sign we need ab=-1

Answer 34

And the equation to be solved
\[
\frac{462 a^5}{b^6}=-\frac{462 a^6}{b^5}
\]

Answer 35

The second equation becomes the first, if we change a=-1/b and b =-1/a
meaning ab=-1

Answer 36

@eliassaab do you have any advice for keeping track of minus signs?

Answer 37

It alternates form + to -1 if you apply the binomial theorem to the power of the
difference.

Answer 38

I meant while working on math questions in general @eliassaab ?

Answer 39

You cannot make a statement about some general question like this one

Answer 40

Thank you