Find the general term for (2x-5y)^7(2x5y)7?

2 Answers
Jul 20, 2018

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

Explanation:

The general (r+1)^(th) term T_(r+1) in the expansion of

(a+b)^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=2x, b=-5y, 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=^7C_r(2x)^{7-r}(-5y)^r

Explanation:

Using binomial expansion of (2x-5y)^7

(2x-5y)^7

=^7C_0(2x)^7+^7C_1(2x)^6(-5y)+ \cdots +^7C_r(2x)^{7-r}(-5y)^r+\ldots +^7C_7(-5y)^7

The general rth term of above Binomial expansion:

T_r=^7C_r(2x)^{7-r}(-5y)^r