chapter
B.2 Two Basic Binomial Theorems
Theorem B.3.
Pages 1

Theorem B.3. For arbitrary x and y, integers m and n (n ≥ m) and arbitrary α and β,

{ n−m k −m

} xk−myn−kQ(

k−m 2 )R(

(xQj + yRj), (B.21)

{ n−m k −m

} xk−myn−kQ(

k−m 2 )+α(k−m)R(

= n−m−1∏

(xQj+α + yRj+β). (B.22)

Proof: It is sufficient to prove (B.21) for m = 0, because the general case follows by substitution of k + m for k and n + m for n. Let

S(n, x, y) = ∑ k

{ n

k

} xkyn−kQ(

k 2)R(

Using { n

k

} = { n− 1 k − 1

} Qn−k +

{ n− 1 k

} Rk ,

we split S into two sums. In the sum with {

we can substitute k + 1 for k. When we take advantage of common factors in the two sums, this gives us

S(n, x, y) = ∑ k

{ n− 1 k

} xkyn−1−kQ(

k 2)R(

n−k 2 )(xQn−1 + yRn−1)

= S(n− 1, x, y)(xQn−1 + yRn−1) .