# What is the fourth term in the expansion of (2x-y)^5?

Feb 13, 2016

You can find this by using the formula ${t}_{r + 1} {=}_{n} {C}_{r} \times {a}^{n - r} \times {b}^{r}$

#### Explanation:

r can be found by solving the simple equation ${t}_{r + 1} = 4$ We can immediately eliminate the t, and we find that r is 3. n is the exponent, which is 5 in this case. Note the n and the r are in base at the C.

${t}_{4} {=}_{5} {C}_{3} \times 2 {x}^{5 - 3} \times - {y}^{3}$

${t}_{4} = 10 \times 4 {x}^{2} \left(- {y}^{3}\right)$

${t}_{4} = - 40 {x}^{2} {y}^{3}$

Your fourth term is $- 40 {x}^{2} {y}^{3}$.

Practice exercises:

1. Find the 7th term in ${\left(- 2 x + 3 y\right)}^{12}$

2. Find the middle term of ${\left(3 x - {y}^{3}\right)}^{8}$

Good luck!