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

User Joshdholtz
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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
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