# Find the general term for (2x-5y)^7?

Jul 20, 2018

 ""_7C_r*128*(-5)^r*x^(7-r)*y^r; r=0,1,2,...,7.

#### Explanation:

The general ${\left(r + 1\right)}^{t h}$ 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, where, r=0,1,2,...,n.

In our case, we have, $a = 2 x , b = - 5 y , n = 7$.

:. T_(r+1)=""_7C_r(2x)^(7-r)(-5y)^r, where, r=0,1,2,...,7,

 i.e., T_(r+1)=""_7C_r*128*(-5)^r*x^(7-r)*y^r; r=0,1,2,...,7#.

${T}_{r} {=}^{7} {C}_{r} {\left(2 x\right)}^{7 - r} {\left(- 5 y\right)}^{r}$

#### Explanation:

Using binomial expansion of ${\left(2 x - 5 y\right)}^{7}$

${\left(2 x - 5 y\right)}^{7}$

${=}^{7} {C}_{0} {\left(2 x\right)}^{7} {+}^{7} {C}_{1} {\left(2 x\right)}^{6} \left(- 5 y\right) + \setminus \cdots {+}^{7} {C}_{r} {\left(2 x\right)}^{7 - r} {\left(- 5 y\right)}^{r} + \setminus \ldots {+}^{7} {C}_{7} {\left(- 5 y\right)}^{7}$

The general $r$th term of above Binomial expansion:

${T}_{r} {=}^{7} {C}_{r} {\left(2 x\right)}^{7 - r} {\left(- 5 y\right)}^{r}$