Find the indicated term of each expansion.
fifth term of (x^2+squ. root 3x)^8
Please show work I need to learn how to do it.
To find the fifth term of the expansion of (x^2 + √(3x))^8, we can use the Binomial Theorem. The Binomial Theorem states that for any positive integer n:
(x + y)^n = C(n,0)*x^n*y^0 + C(n,1)*x^(n-1)*y^1 + C(n,2)*x^(n-2)*y^2 + ... + C(n,n)*x^0*y^n
where C(n,r) represents the binomial coefficient, given by:
C(n,r) = n! / (r!(n-r)!)
In our case, x^2 is equivalent to x^2 * 1, and we can consider √(3x) as another variable, let’s say y. Therefore, we have:
(x^2 + √(3x))^8 = C(8,0)*x^(2*8)*y^0 + C(8,1)*x^(2*8 - 1)*y^1 + C(8,2)*x^(2*8 - 2)*y^2 + ... + C(8,8)*x^0*y^(8*1)
To find the fifth term, we need to consider the term with x raised to the (8 - 5)th power and y raised to the 5th power. Let's calculate this term:
Term 5 = C(8,5) * x^(2*8 - 5) * y^5
C(8,5) = 8! / (5!(8-5)!) = (8 * 7 * 6) / (3 * 2 * 1) = 56
Therefore, the fifth term of the expansion is:
Term 5 = 56 * x^(16 - 5) * y^5 = 56 * x^11 * y^5
In this case, y represents √(3x), so the final answer is:
Term 5 = 56 * x^11 * (√(3x))^5 = 56 * x^11 * (3x)^(5/2)
You can simplify this expression further if needed.