# How do you find the fourth term of (x+2)^7?

Jan 11, 2017

$280 {x}^{4}$

#### Explanation:

Recall that the ${\left(r + 1\right)}^{t h}$ term ${T}_{r + 1}$ in the expansion of

${\left(a + b\right)}^{n}$ is given by, T_(r+1)=""_nC_ra^(n-r)b^r.

In our Example, we have,

n=7, a=x, b=2, &, r+1=4, i.e., r=3

:. T_(3+1)=T_4=""_7C_3(x^(7-3))(2^3)=(7.6.5)/(1.2.3)(x^4)8

Hence, the reqd. term ${T}_{4} = 280 {x}^{4}$.

Alternatively, we can expand ${\left(x + 2\right)}^{7}$ using the Binomial

Theorem ** upto the reqd. ${4}^{t h}$ term, as shown below :

${\left(x + 2\right)}^{7} = \text{_7C_0(x^(7-0))(2^0)+""_7C_1(x^(7-1))(2^1)+""_7C_2(x^(7-2))(2^2)+""_7C_3(x^(7-3))(2^3)+...+"Last Term}$, giving the same Answer.