# Binomial expansion

• Apr 17th 2010, 06:21 AM
Punch
Binomial expansion
Find the coefficients of $\displaystyle x^7$ in the expansion of $\displaystyle (1-3x^2)(x^2-\frac{2}{x})^7$

My attempt:

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

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

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

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

Puzzled at arriving at $\displaystyle 2\frac{1}{3}$ for $\displaystyle r$.
• Apr 17th 2010, 06:39 AM
Quote:

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

My attempt:

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

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

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

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

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

is the answer 840 ?
• Apr 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
• Apr 17th 2010, 09:35 PM
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 :

$\displaystyle (a+b)^n=\sum^{n}_{k=0} \binom{n}{k} a^{n-k}b^k$

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

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

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

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 $\displaystyle \binom{7}{3} \cdot 3 \cdot 2^3=840$
• Apr 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?
• Apr 18th 2010, 01:45 AM
Quote:

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

My attempt:

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

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

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

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

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

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

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

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

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

check your 2nd last line simplification again

.
• Apr 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
• Apr 18th 2010, 06:33 AM
Yes, it boils down to finding the $\displaystyle x^5$ term of the binomial.
Then the coefficient of $\displaystyle x^7$ is $\displaystyle -3(coefficient\ of\ x^5\ from\ the\ binomial\ expansion)$