Question

Find the first 3 and last 3 terms in the expansion `(2x-1)^11` using the binomial theorem?

Answer 1

`-1,22x,-220x^2,28160x^9,-11264x^10,2048x^11`

Explanation:

`(ax+b)^n=sum_(r=0)^n((n),(r))(ax)^rb^(n-r)=sum_(r=0)^n(n!)/(r!(n-r)!)(ax)^rb^(n-r)`

So, we want `rin{0,1,2,9,10,11}`

`(11!)/(0!(11-0)!)(2x)^0(-1)^11=1(1)(-1)=-1`
`(11!)/(1!(11-1)!)(2x)^1(-1)^10=11(2x)(1)=22x`
`(11!)/(2!(11-2)!)(2x)^2(-1)^9=55(4x^2)(-1)=-220x^2`
`(11!)/(9!(11-9)!)(2x)^9(-1)^2=55(512x^9)(1)=28160x^9`
`(11!)/(10!(11-10)!)(2x)^10(-1)^1=11(1024x^10)(-1)=-11264x^10`
`(11!)/(11!(11-11)!)(2x)^11(-1)^0=1(2048x^11)(1)=2048x^11`

These are the first 3 and last 3 terms in order of increasing powers of `x`:
`-1,22x,-220x^2,28160x^9,-11264x^10,2048x^11`