# How do you find the fifth term of (2a+3b)^10?

Feb 16, 2017

$1088640 {a}^{6} {b}^{4}$

#### Explanation:

In the binomial expansion of ${\left(x + a\right)}^{n}$, (r+1)th term is written as,

${T}_{r + 1} {=}^{n} {C}_{r} {x}^{n - r} {a}^{r}$. Here n=10, x=2a and b=3b

For 5th term, put r=4 in this formula

${T}_{5} {=}^{10} {C}_{4} {\left(2 a\right)}^{6} {\left(3 b\right)}^{4}$

= $\frac{10.9 .8 .7}{1.2 .3 .4} \left(64 {a}^{6}\right) \left(81 {b}^{4}\right)$

=$1088640 {a}^{6} {b}^{4}$