in the expansion of (x^2+3)^10 what is the coefficient of the term containing x^16. I did 10c4 * 3. But i am not getting the correct answer. can someone please show me how to do this. thanks for your time!
the general term is
GT(n+1) = C(10,n)(x^2)^n (3)^(10-n)
= C(10,n)(x)^2n (3)^(10-n)
so the x^16 has to come from the x^2n
and 2n = 16
n = 8
so it is term(9) and it is
C(10,9)(x^2)^8 (3)^(10-8)
= 10(x^16)(9)
= 90x^16
so the coefficient is 90
<<
so it is term(9) and it is
C(10,9)(x^2)^8 (3)^(10-8)
= 10(x^16)(9)
= 90x^16
so the coefficient is 90 >>
should have said:
so it is term(9) and it is
C(10,8)(x^2)^8 (3)^(10-8)
= 45(x^16)(9)
= 405x^16
so the coefficient is 405
thanks!
To find the coefficient of the term containing x^16 in the expansion of (x^2 + 3)^10, you need to use the Binomial Theorem. The Binomial Theorem states that for any positive integer n, the expansion of (a + b)^n can be written as the sum of the terms of the form C(n, k) * a^(n-k) * b^k, where C(n, k) represents the binomial coefficient and is calculated using the formula:
C(n, k) = n! / (k! * (n - k)!)
In your case, the expression (x^2 + 3)^10 suggests that a = x^2 and b = 3, with n = 10. To find the coefficient of the term containing x^16, we need to determine the value of k in the expression C(10, k) * (x^2)^(10-k) * 3^k, where the exponent of x is 16. So, we can set up the equation:
10 - k = 16
Solving this equation gives us k = -6, but since k cannot be negative, there is no term with x^16 in the expansion (x^2 + 3)^10. Therefore, the coefficient is 0.
In summary, the coefficient of the term containing x^16 in the expansion of (x^2 + 3)^10 is 0.