# Question ae7e8

Mar 17, 2017

The coefficient of ${x}^{2}$ in the expansion of:

$P \left(x\right) = {\left(1 + 3 x\right)}^{6} \left(1 - 3 x - 5 {x}^{2}\right)$

is: ${c}_{2} = 76$

#### Explanation:

We have:

$P \left(x\right) = {\sum}_{n = 0}^{6} {c}_{n} {x}^{n} = {\left(1 + 3 x\right)}^{6} \left(1 - 3 x - 5 {x}^{2}\right)$

Develop ${\left(1 + 3 x\right)}^{6}$ using binomial coefficients:

${\sum}_{n = 0}^{6} {c}_{n} {x}^{n} = \left(1 - 3 x - 5 {x}^{2}\right) {\sum}_{n = 0}^{6} \left(\begin{matrix}6 \\ n\end{matrix}\right) {\left(3 x\right)}^{n}$

Now using the distributive property of multiplication:

${\sum}_{n = 0}^{6} {c}_{n} {x}^{n} = {\sum}_{n = 0}^{6} \left(\begin{matrix}6 \\ n\end{matrix}\right) {3}^{n} {x}^{n} + {\sum}_{n = 0}^{6} \left(\begin{matrix}6 \\ n\end{matrix}\right) \left(- 3 x\right) {3}^{n} {x}^{n} + {\sum}_{n = 0}^{6} \left(\begin{matrix}6 \\ n\end{matrix}\right) \left(- 5 {x}^{2}\right) {3}^{n} {x}^{n}$

${\sum}_{n = 0}^{6} {c}_{n} {x}^{n} = {\sum}_{n = 0}^{6} \left(\begin{matrix}6 \\ n\end{matrix}\right) {3}^{n} {x}^{n} - {\sum}_{n = 0}^{6} \left(\begin{matrix}6 \\ n\end{matrix}\right) {3}^{n + 1} {x}^{n + 1} - 5 {\sum}_{n = 0}^{6} \left(\begin{matrix}6 \\ n\end{matrix}\right) {3}^{n} {x}^{n + 2}$

The coefficient of ${x}^{2}$ is then the sum of the term for $n = 2$ in the first sum, for $n = 1$ in the second sum and for $n = 0$ in the third sum:

${c}_{2} = \left(\begin{matrix}6 \\ 2\end{matrix}\right) \cdot {3}^{2} - \left(\begin{matrix}6 \\ 1\end{matrix}\right) \cdot {3}^{2} - 5 \cdot \left(\begin{matrix}6 \\ 0\end{matrix}\right)$

The general expression of the binomial coefficient is:

((n),(k)) = (n!)/(k!(n-k)!)

So:

((6),(2)) = (6!)/((2!)(4!)) = 15

((6),(1)) = (6!)/((1!)(5!)) = 6

((6),(0)) = (6!)/((0!)(6!)) = 1

and:

${c}_{2} = 15 \cdot 9 - 6 \cdot 9 - 5 = 76$

Mar 17, 2017

$76.$

#### Explanation:

Knowing that, ${\left(1 + 3 x\right)}^{6}$

$= 1 + {\text{_6C_1(3x)+""_6C_2(3x)^2+...+}}_{6} {C}_{6} {\left(3 x\right)}^{6.}$

$= 1 + \left(6\right) \left(3 x\right) + \frac{\left(6\right) \left(5\right)}{\left(1\right) \left(2\right)} \left(9 {x}^{2}\right) + \ldots + 729 {x}^{6.}$

$= 1 + 18 x + 135 {x}^{2} + \ldots + 729 {x}^{6.}$

$\therefore {\left(1 + 3 x\right)}^{6} \left(1 - 3 x - 5 {x}^{2}\right)$

$= \left(1 + 18 x + 135 {x}^{2} + \ldots + 729 {x}^{6}\right) \left(1 - 3 x - 5 {x}^{2}\right)$

$= \left(1 + 18 x + 135 {x}^{2} + \ldots\right) - 3 x \left(1 + 18 x + 135 {x}^{2.} . .\right) - 5 {x}^{2} \left(1 + 18 x + 135 {x}^{2} + \ldots\right) + \ldots$

$= 1 + 18 x + 135 {x}^{2} + \ldots - 3 x - 54 {x}^{2} - 405 {x}^{3.} . . - 5 {x}^{2} - 90 {x}^{3} - 675 {x}^{4.} . .$

=1+15x+76x^2+...

$\therefore \text{ The Desired Co-eff. of "x^2" is } 76.$

Enjoy Maths.!