Problem 59 - Beginning and Intermediate Algebra

Question

Find the indicated term of each binomial expansion. $$\left(2 y^{2}+z\right)^{10} ; \text { eighth term }$$

Step-by-Step Solution

Answer

The eighth term in the expansion of the given binomial expression, \((2y^2+z)^{10}\), is \(T_8 = 960y^{6}z^{7}\).

Step 1: Identify the components of the binomial expression

In this problem, we have the given binomial expression \((2y^2+z)^{10}\) and we need to find the eighth term in its expansion. The components of the binomial expression are: - a = \(2y^2\) - b = z - n = 10, which is the exponent.

Step 2: Understand the general term of a binomial expansion

The general term of a binomial expansion according to the binomial theorem is given by: \(T_r = C(n, k) * a^{n-k}*b^k\), where: - \(T_r\) is the r-th term in the expansion, - \(C(n, k)\) is the binomial coefficient (also denoted as \(\binom{n}{k}\) or nCk), - n is the exponent in the binomial expression, - a and b are the items of the binomial expression, and - k ranges from 0 to n. Since we need to find the eighth term, \(T_8\), we will use \(k = 7\) (as the first term corresponds to \(k = 0\)).

Step 3: Calculate the binomial coefficient

To find the binomial coefficient for the eighth term, we need to calculate: \(C(10, 7) = \binom{10}{7} = \frac{10!}{7!(3)!}\), where ! denotes the factorial of a number. We can calculate this as: \(C(10, 7)= \frac{10!}{7!3!} = \frac{10\times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120\).

Step 4: Apply the formula for the general term

Now, we can apply the formula for the eighth term, \(T_8\), using \(k = 7\): \(T_8 = C(10,7) * (2y^2)^{10-7} * z^7\) Plug in the binomial coefficient we found in step 3: \(T_8 = 120*(2y^2)^3*z^7\)

Step 5: Simplify the term

Now, simplify the term: \(T_8 = 120*(8y^6)*z^7\) \(T_8 = 960y^{6}z^{7}\) So, the eighth term in the expansion of the given binomial expression is: \(T_8 = 960y^{6}z^{7}\)