Binomial expansion

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April 17th 2010, 06:21 AM
Punch

Binomial expansion

Find the coefficients of $x^7$ in the expansion of $(1-3x^2)(x^2-\frac{2}{x})^7$

My attempt:

to obtain $x^7$ , coefficients of $x^5$ and $x^7$ have to be used.

$T_{r+1}= (7Cr)(x^2)^{7-r}(\frac{-2}{x})^r$

therefore, $2(7-r)+(-r)=5$
$r=3$

OR $2(7-r)+(-r)=5$
$r= 2(\frac{1}{3})$

Puzzled at arriving at $2\frac{1}{3}$ for $r$ .
April 17th 2010, 06:39 AM
mathaddict

Quote:

Originally Posted by Punch

Find the coefficients of $x^7$ in the expansion of $(1-3x^2)(x^2-\frac{2}{x})^7$

My attempt:

to obtain $x^7$ , coefficients of $x^5$ and $x^7$ have to be used.

$T_{r+1}= (7Cr)(x^2)^{7-r}(\frac{-2}{x})^r$

therefore, $2(7-r)+(-r)=5$
$r=3$

OR $2(7-r)+(-r)=5$
$r= 2(\frac{1}{3})$

Puzzled at arriving at $2\frac{1}{3}$ for $r$ .

is the answer 840 ?
April 17th 2010, 05:43 PM
Punch

Quote:

Originally Posted by mathaddict

is the answer 840 ?

yes it is! please show me the right way of solving it
April 17th 2010, 09:35 PM
mathaddict

Quote:

Originally Posted by Punch

yes it is! please show me the right way of solving it

Great . Of course , we can expand the whole thing and combine terms which will produce x^7 but it would be tedious and risky so its better to use the binomial theorem :

$<br /> (a+b)^n=\sum^{n}_{k=0} \binom{n}{k} a^{n-k}b^k<br />$

$<br /> (x^2-\frac{2}{x})^7=\sum^{n}_{k=0} \binom{7}{k}x^{14-3k}\cdot 2^k<br />$

$(1-3x^2)(x^2-\frac{2}{x})^7$

$=\sum^{n}_{k=0} \binom{7}{k}x^{14-3k}\cdot 2^k+\sum^{n}_{k=0}\binom{7}{k} x^{16-3k}\cdot 3\cdot 2^k$

so now lets see what values of k give 7

14-3k=7 , k=7/3 and this cant be since k must be +ve integer

how about 16-3k=7 , k=3 . (Nod)

its coefficient is $\binom{7}{3} \cdot 3 \cdot 2^3=840$
April 18th 2010, 01:12 AM
Punch

I didnt quite understand "so now lets see what values of k give 7

14-3k=7 , k=7/3 and this cant be since k must be +ve integer

how about 16-3k=7 , k=3 ."

Does this shows that if the value of r is not a whole number, we have to round the number off and see if the power of x is what we are after?
April 18th 2010, 01:45 AM
Archie Meade

Quote:

Originally Posted by Punch

Find the coefficients of $x^7$ in the expansion of $(1-3x^2)(x^2-\frac{2}{x})^7$

My attempt:

to obtain $x^7$ , coefficients of $x^5$ and $x^7$ have to be used.

$\color{blue}x^0$ and $\color{blue}x^7$ , $\color{blue}x^2$ and $\color{blue}x^5$

$T_{r+1}= (7Cr)(x^2)^{7-r}(\frac{-2}{x})^r$

the above is fine if you recognise that there is no $\color{blue}x^7$ term when you multiply the binomial expansion by 1

$\color{blue}x^{14-2r}x^{-r}=x^{14-3r}=x^7$ has no natural solution

Then the $\color{blue}x^7$ term is obtained by multiplying the binomial expansion by $\color{blue}-3x^2$

therefore, $2(7-r)+(-r)=5$
$r=3$

OR $2(7-r)+(-r)=5$
$r= 2(\frac{1}{3})$ you should have 14-2r-r=14-3r=5, 3r=14-5=9

Puzzled at arriving at $2\frac{1}{3}$ for $r$ .

check your 2nd last line simplification again

.
April 18th 2010, 04:53 AM
Punch

alright so can i conclude that I only have to use the term coefficient of x^5 since there is no x^7 term
April 18th 2010, 06:33 AM
Archie Meade

Quote:

Originally Posted by Punch

alright so can i conclude that I only have to use the term coefficient of x^5 since there is no x^7 term

Yes, it boils down to finding the $x^5$ term of the binomial.

Then the coefficient of $x^7$ is $-3(coefficient\ of\ x^5\ from\ the\ binomial\ expansion)$

This is a shortcut to multiplying out all the terms independently,
if you only want a single term of the entire expansion of the factors.