Answer:
Part A)
[tex]\displaystyle z_6 = 9609600000x^{10}[/tex]
Part B)
[tex]z_4 = 10500 x^9[/tex]
Step-by-step explanation:
Recall the binomial expansion theorem:
[tex]\displaystyle (x+y)^n = \sum_{k=0}^{n}{n \choose k} x^{n-k} y^k[/tex]
Part A)
Our expression is equivalent to:
[tex]\displaystyle (2x+5)^{15} = \sum_{k = 0}^{15} {15 \choose k} (2x)^{15-k}\cdot 5^k[/tex]
To find the sixth term, let k = 5. Therefore, the sixth term is:
[tex]\displaystyle \begin{aligned} z_6 &= {15\choose 5} (2x)^{15-5}\cdot 5^5 \\ \\ & = {15\choose 5}x^{10} \cdot (2)^{10}\cdot 5^5 \\ \\ &= 9609600000x^{10}\end{aligned}[/tex]
Part B)
Likewise:
[tex]\displaystyle \begin{aligned} \left(x^2 + \frac{5}{x}\right)^9 = \sum_{k=0}^9 {9\choose k}(x^2)^{9-k}\left(\frac{5}{x}\right)^{k}\end{aligned}[/tex]
To find the fourth term, let k = 3. Therefore, the fourth term is:
[tex]\displaystyle \begin{aligned} z_4 & = {9\choose 3}\left(x^2\right)^{9-3} \left(\frac{5}{x}\right)^{3} \\ \\ & = {9\choose 3}x^{12} \cdot \frac{5^3}{x^3} \\ \\ & = 10500x^9\end{aligned}[/tex]
