Coefficient of x^4 in the product of the expansions (1 + x)^5.(1 - x)^6 is -

A) 5C2

B) ∑(r = 0 to 4) (-1)^r 6Cr . 5C(4 - r)

C) - 5C3

D) 5C4

More than one options may be correct.

- Feb 13th 2009, 10:27 PMfardeen_genCoefficient of x^4 in the product of the expansions (1 + x)^5.(1 - x)^6 is?
Coefficient of x^4 in the product of the expansions (1 + x)^5.(1 - x)^6 is -

A) 5C2

B) ∑(r = 0 to 4) (-1)^r 6Cr . 5C(4 - r)

C) - 5C3

D) 5C4

More than one options may be correct. - Feb 13th 2009, 10:50 PMred_dog
The general term of $\displaystyle (1+x)^5$ is $\displaystyle C_5^kx^k, \ 0\leq k\leq 5$

The general term of $\displaystyle (1-x)^6$ is $\displaystyle (-1)^lC_6^lx^l, \ 0\leq l\leq 6$

Then the genaral term of the entire expansion is $\displaystyle (-1)^lC_5^kC_6^lx^{k+l}, \ 0\leq k\leq 5, \ 0\leq l\leq 6$

So the condition is $\displaystyle k+l=4$.

Now take all the possibilities for k and l. - Feb 13th 2009, 10:55 PMDeMath
Hint:

$\displaystyle {\left( {1 + x} \right)^5}{\left( {1 - x} \right)^6} = {\left( {1 + x} \right)^5}{\left( {1 - x} \right)^5}\left( {1 - x} \right) = {\left( {1 - {x^2}} \right)^5}\left( {1 - x} \right) = {\left( {{x^2} - 1} \right)^5}\left( {x - 1} \right)$ - Feb 13th 2009, 11:08 PMfardeen_genQuote:

The general term of http://www.mathhelpforum.com/math-he...28391dc6-1.gif is http://www.mathhelpforum.com/math-he...53b23532-1.gif

The general term of http://www.mathhelpforum.com/math-he...021d471d-1.gif is http://www.mathhelpforum.com/math-he...7202a132-1.gif

Then the genaral term of the entire expansion is http://www.mathhelpforum.com/math-he...d22b7052-1.gif

So the condition is http://www.mathhelpforum.com/math-he...9059a502-1.gif.

Now take all the possibilities for k and l.

- Feb 13th 2009, 11:23 PMred_dog
No.

You have the possibilities:

k=0, l=4

k=1, l=3

k=2, l=2

k=3, l=1

k=4, l=0

Then the coefficient is

$\displaystyle C_5^0C_6^4-C_5^1C_6^3+C_5^2C_6^2-C_5^3C_6^1+C_5^4C_6^0=\sum_{r=0}^4(-1)^rC_6^rC_5^{4-r}$ - Feb 13th 2009, 11:29 PMDeMath
Members of the decomposition of the factor $\displaystyle {\left( {{x^2} - 1} \right)^5}$ have only even powers. Therefore, you just enough to consider this factor.

Your answer must be $\displaystyle C_5^3 = 10$ for $\displaystyle x^4$. - Feb 13th 2009, 11:37 PMfardeen_gen
So correct answers are:

B), C) and D), right? - Feb 13th 2009, 11:43 PMred_dog
No, the correct answer is B.

- Feb 13th 2009, 11:47 PMDeMath
$\displaystyle {\left( {{x^2} - 1} \right)^5} = \sum\limits_{k = 0}^5 {C_5^k{x^{2\left( {5 - k} \right)}}{{\left( { - 1} \right)}^k}} = C_5^0{x^{10}} - C_5^1{x^8} + C_5^2{x^6} - C_5^3{x^4} + C_5^4{x^2} - C_5^5{x^0}$

$\displaystyle {\left( {1 + x} \right)^5}{\left( {1 - x} \right)^6} = {\left( {{x^2} - 1} \right)^5}\left( {x - 1} \right) \Rightarrow C_5^3 = 10{\text{ for }}{x^4}$

Sorry, I corrected. - Feb 13th 2009, 11:53 PMred_dog
- Feb 13th 2009, 11:56 PMfardeen_gen
red_dog, A) is equivalent to B). So A) is also correct, right?

And aren't we supposed to take the individual possibility of k = 4 and l = 0, which gives 5C4(option D)?