How do you use the binomial series to expand # (2+kx)^8 #?

1 Answer
Jan 21, 2017

Answer:

See the answer below

Explanation:

The binomial theorem is

#(a+b)^n=sum_(r=0)^n((n),(r))a^(n-r)b^r#

#((n),(r))=(n!)/(r!(n-r)!)#

Therefore,

#(2+kx)^8=sum_(r=0)^8((8),(r))2^(8-r)(kx)^r#

#=((8),(0))2^8+((8),(1))2^7(kx)+((8),(2))2^6(kx)^2+((8),(3))2^5(kx)^3+((8),(4))2^4(kx)^4+((8),(5))2^3(kx)^5+((8),(6))2^2(kx)^6+((8),(7))2(kx)^7+((8),(8))2^0(kx)^8#

#=256+1024kx+1792k^2x^2+1762k^3x^3+1120k^4x^4+448k^5x^5+112k^6x^6+14k^7x^7+k^8x^8#