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

**fardeen_gen** · February 13th 2009, 10:27 PM

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.

**red_dog** · February 13th 2009, 10:50 PM

The general term of $(1+x)^5$ is $C_5^kx^k, \ 0\leq k\leq 5$

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

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

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

Now take all the possibilities for k and l.

**DeMath** · February 13th 2009, 10:55 PM

Originally Posted by fardeen_gen

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.

Hint:

${\left( {1 + x} \right)^5}{\left( {1 - x} \right)^6} = {\left( {1 + x} \right)^5}{\left( {1 - x} \right)^5}\left( {1 - x} \right) = {\left( {1 - {x^2}} \right)^5}\left( {1 - x} \right) = {\left( {{x^2} - 1} \right)^5}\left( {x - 1} \right)$

**fardeen_gen** · February 13th 2009, 11:08 PM

The general term of

is

The general term of

is

Then the genaral term of the entire expansion is

So the condition is

.

Now take all the possibilities for k and l.

Correct me if I am wrong. 5C4 is the only possible option.

**red_dog** · February 13th 2009, 11:23 PM

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

$C_5^0C_6^4-C_5^1C_6^3+C_5^2C_6^2-C_5^3C_6^1+C_5^4C_6^0=\sum_{r=0}^4(-1)^rC_6^rC_5^{4-r}$

**DeMath** · February 13th 2009, 11:29 PM

Members of the decomposition of the factor ${\left( {{x^2} - 1} \right)^5}$ have only even powers. Therefore, you just enough to consider this factor.

Your answer must be $C_5^3 = 10$ for $x^4$ .

**fardeen_gen** · February 13th 2009, 11:37 PM

So correct answers are:

B), C) and D), right?

**red_dog** · February 13th 2009, 11:43 PM

No, the correct answer is B.

**DeMath** · February 13th 2009, 11:47 PM

${\left( {{x^2} - 1} \right)^5} = \sum\limits_{k = 0}^5 {C_5^k{x^{2\left( {5 - k} \right)}}{{\left( { - 1} \right)}^k}} = C_5^0{x^{10}} - C_5^1{x^8} + C_5^2{x^6} - C_5^3{x^4} + C_5^4{x^2} - C_5^5{x^0}$

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

Sorry, I corrected.

**red_dog** · February 13th 2009, 11:53 PM

Originally Posted by DeMath

Only C)

${\left( {{x^2} - 1} \right)^5} = \sum\limits_{k = 0}^5 {C_5^k{x^{2\left( {5 - k} \right)}}{{\left( { - 1} \right)}^k}} = C_5^0{x^{10}} - C_5^1{x^8} + C_5^2{x^6} - C_5^3{x^4} + C_5^4{x^2} - C_5^5{x^0}$

$(1+x)^5(1-x)^6\neq (x^2-1)^5$ !!!

**fardeen_gen** · February 13th 2009, 11:56 PM

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)?

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