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

Question

2709 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-20T18:49:56

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

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#.

Answer 2 · 2018-07-20T19:40:45

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

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 #r#th term of above Binomial expansion:

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