How do you use the binomial series to expand (1+x)^n(1+x)n?

1 Answer
sente
Dec 10, 2015

Assuming nn is a nonnegative integer, then the binomial theorem states that

(a+b)^n = sum_(k=0)^nC(n,k)a^(n-k)b^k= sum_(k=0)^n(n!)/(k!(n-k)!)a^(n-k)b^k(a+b)n=n∑k=0C(n,k)an−kbk=n∑k=0n!k!(n−k)!an−kbk

Applying it in this case with a = 1a=1 and b = xb=x, we get

(1+x)^n = sum_(k=0)^n(n!)/(k!(n-k)!)1^(n-k)x^k = sum_(k=0)^n(n!)/(k!(n-k)!)x^k(1+x)n=n∑k=0n!k!(n−k)!1n−kxk=n∑k=0n!k!(n−k)!xk

= x^n + nx +(n(n-1))/2x^2+(n(n-1)(n-2))/6x^3+...+(n(n-1))/2x^(n-2) + nx^(n-1) + 1

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