MacLaurin Series From Binomial Series Formula

Question

Use the following binomial series formula:

(1+x)^m = 1 + mx + [m(m−1)x^2]/2! + ⋅⋅⋅ + [m(m−1)⋯(m−k+1)x^k]/k! + ⋅⋅⋅

to obtain the MacLaurin series for

1/[(1+x)^9] = ∑k=0 to ∞ _______

Answer 1

Observe 1/(1 + x)⁹ = (1 + x)^-9. Using the binomial series formula with m = -9 we obtain

1/(1 + x)⁹ = ∑[k = 0 to ∞] binom{-9, k} x^k

where binom{-9, k} = (-9)(-9-1)•••(-9-k+1)/k! Now

(-9)(-9-1)•••(-9-k+1)/k! = (-1)^k (9+k-1)•••(9+1)(9)/k! = (-1)^k (9 + k - 1 choose k) = (-1)^k (8 + k choose k)

and hence

1/(1 + x)⁹ = ∑[k = 0 to ∞] (-1)^k (8 + k choose k) x^k = ∑[k = 0 to ∞] (-1)^k (8 + k)!/(8! k!) x^k