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.
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.
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.
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 $\displaystyle x^4$.
$\displaystyle {\left( {{x^2} - 1} \right)^5} = \sum\limits_{k = 0}^5 {C_5^k{x^{2\left( {5 - k} \right)}}{{\left( { - 1} \right)}^k}} = C_5^0{x^{10}} - C_5^1{x^8} + C_5^2{x^6} - C_5^3{x^4} + C_5^4{x^2} - C_5^5{x^0}$
$\displaystyle {\left( {1 + x} \right)^5}{\left( {1 - x} \right)^6} = {\left( {{x^2} - 1} \right)^5}\left( {x - 1} \right) \Rightarrow C_5^3 = 10{\text{ for }}{x^4}$
Sorry, I corrected.