Question #ae7e8

2 Answers
Mar 17, 2017

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

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

is: #c_2=76#

Explanation:

We have:

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

Develop #(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.!