190k views
2 votes
Find the value of a of coefficient of x² in the expansion (1+ax)⁴(a-2)³ is 6​

User Joshdholtz
by
8.6k points

1 Answer

6 votes

Explanation:

To find the coefficient of x² in the expansion of (1+ax)⁴(a-2)³, we can use the binomial theorem. The binomial theorem states that for any two numbers a and b and any non-negative integer n, the expansion of (a+b)^n is given by the sum of the terms:

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

where C(n,k) = n! / (k!(n-k)!) is the binomial coefficient.

We can apply the binomial theorem to find the coefficient of x² in the expansion of (1+ax)⁴(a-2)³:

(1+ax)⁴(a-2)³ = (1+4ax+6(ax)²+4(ax)³+(ax)⁴)(a-2)³ = a^3 - 6a^2 + 12a - 8 + 6(ax)² + ...

So the coefficient of x² in the expansion is 6.

User THeSiD
by
8.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories