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Find the coefficient of the third term of (x+2)^5

User ZephDavies
by
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1 Answer

6 votes

Answer:

40

Explanation:

(x+2)^5 use binomial theorem :

(a+b)^n = (n choose 0)*a^n*b^0 + (n choose 1)*a^(n-1)*b^1 + (n choose 2)*a^(n-2)*b^2) + ... + (n choose (n-1)*a^1*b^(n-1) + ( n choose n)*a^0*b^n

this seems like a lot but to break it down, notice how the exponent on 'a' decreases as the exponent on 'b' gets bigger.

also, the 'choose' formula is :

(n choose r ) = n!/ (n-r)!r!

now plug in your values

(x+2)^5 =

(5 choose 0)*x^5*2^0 + (5choose 1)*x^4*2^1 + (5 choose 2)*x^3*x^2 + (5 choose 3)*x^2*2^3 + (5 choose 4)*x^1*2^4 + (5 choose 5)*x^0*x^5

we only need the third term so we will solve for this :

(5 choose 2)*x^3*x^2

5 choose 2 = 5!/ (5-2)!2! = 5!/ 3!2! = 10

x^3 * 2^2 = 4x^3

10*4x^3 = 40x^3

User Nifoem Bar
by
4.7k points