# How do you find the 7th term in the expansion of the binomial (7x+2y)^15?

Aug 18, 2017

$7$ th term is $15 {C}_{6} \cdot {\left(7 x\right)}^{9} \cdot {\left(2 y\right)}^{6} = 1.29E+13 \cdot {x}^{9} \cdot {y}^{6}$

#### Explanation:

Binomial theorem:
${\left(a + b\right)}^{n} = \left(n {C}_{0}\right) {a}^{n} {b}^{0} + \left(n {C}_{1}\right) {a}^{n - 1} {b}^{1} + \left(n {C}_{2}\right) {a}^{n - 2} {b}^{2} + \ldots \ldots \left(n {C}_{n}\right) {b}^{n}$
Here  (a=7x ; b=2y , n= 15)

We know nC_r= (n!)/(r!(n-r)!) :. 15C_6=5005

$7$ th term is $15 {C}_{6} \cdot {\left(7 x\right)}^{9} \cdot {\left(2 y\right)}^{6} = 1.29E+13 \cdot {x}^{9} \cdot {y}^{6}$ [Ans]