Question #ae7e8

2 Answers
Mar 17, 2017

The coefficient of x^2x2 in the expansion of:

P(x) = (1+3x)^6(1-3x-5x^2)P(x)=(1+3x)6(13x5x2)

is: c_2=76c2=76

Explanation:

We have:

P(x) = sum_(n=0)^6 c_nx^n = (1+3x)^6(1-3x-5x^2)P(x)=6n=0cnxn=(1+3x)6(13x5x2)

Develop (1+3x)^6(1+3x)6 using binomial coefficients:

sum_(n=0)^6 c_nx^n = (1-3x-5x^2)sum_(n=0)^6 ((6),(n))(3x)^n

Now using the distributive property of multiplication:

sum_(n=0)^6 c_nx^n = sum_(n=0)^6 ((6),(n))3^nx^n + sum_(n=0)^6 ((6),(n))(-3x)3^nx^n +sum_(n=0)^6 ((6),(n))(-5x^2)3^nx^n

sum_(n=0)^6 c_nx^n = sum_(n=0)^6 ((6),(n))3^nx^n - sum_(n=0)^6 ((6),(n))3^(n+1)x^(n+1) -5sum_(n=0)^6 ((6),(n))3^nx^(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 = ((6),(2)) * 3^2 - ((6),(1)) * 3^2 -5*((6),(0))

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 * 9 - 6 * 9 -5 = 76

Mar 17, 2017

76.

Explanation:

Knowing that, (1+3x)^6

=1+""_6C_1(3x)+""_6C_2(3x)^2+...+""_6C_6(3x)^6.

=1+(6)(3x)+((6)(5))/((1)(2))(9x^2)+...+729x^6.

=1+18x+135x^2+...+729x^6.

:. (1+3x)^6(1-3x-5x^2)

=(1+18x+135x^2+...+729x^6)(1-3x-5x^2)

=(1+18x+135x^2+...)-3x(1+18x+135x^2...)-5x^2(1+18x+135x^2+...)+...

=1+18x+135x^2+...-3x-54x^2-405x^3...-5x^2-90x^3-675x^4...

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

:." The Desired Co-eff. of "x^2" is "76.

Enjoy Maths.!