stylistic:tone, quality:corruption
Example 2: Modified binomial coefficient sums and the binomial transform

As another example of using generating functions to relate sequences and manipulate sums, for an arbitrary sequence ⟨ f ⟩ we define the two sequences of sums s n:= ∑ m = 0 n (n m) f m 3 n − m s ~ n:= ∑ m = 0 n (n m) (m + 1) (m + 2) (m + 3) f m 3 n − m, {\displaystyle {\begin{aligned}s_{n}&:=\sum _{m=0}^{n}{\binom {n}{m}}f_{m}3^{n-m}\\{\tilde {s}}_{n}&:=\sum _{m=0}^{n}{\binom {n}{m}}(m+1)(m+2)(m+3)f_{m}3^{n-m}\,,\end{aligned}}} for all n ≥ 0, and seek to express the second sums in terms of the first. We suggest an approach by generating functions.

First, we use the binomial transform to write the generating function for the first sum as S (z) = 1 1 − 3 z F (z 1 − 3 z). {\displaystyle S(z)={\frac {1}{1-3z}}F\left({\frac {z}{1-3z}}\right).}

Since the generating function for the sequence ⟨ (n + 1)(n + 2)(n + 3) f ⟩ is given by 6 F (z) + 18 z F ′ (z) + 9 z 2 F ″ (z) + z 3 F ‴ (z) {\displaystyle 6F(z)+18zF'(z)+9z^{2}F''(z)+z^{3}F'''(z)} we may write the generating function for the second sum defined above in the form S ~ (z) = 6 (1 − 3 z) F (z 1 − 3 z) + 18 z (1 − 3 z) 2 F ′ (z 1 − 3 z) + 9 z 2 (1 − 3 z) 3 F ″ (z 1 − 3 z) + z 3 (1 − 3 z) 4 F ‴ (z 1 − 3 z). {\displaystyle {\tilde {S}}(z)={\frac {6}{(1-3z)}}F\left({\frac {z}{1-3z}}\right)+{\frac {18z}{(1-3z)^{2}}}F'\left({\frac {z}{1-3z}}\right)+{\frac {9z^{2}}{(1-3z)^{3}}}F''\left({\frac {z}{1-3z}}\right)+{\frac {z^{3}}{(1-3z)^{4}}}F'''\left({\frac {z}{1-3z}}\right).}

In particular, we may write this modified sum generating function in the form of a (z) ⋅ S (z) + b (z) ⋅ z S ′ (z) + c (z) ⋅ z 2 S ″ (z) + d (z) ⋅ z 3 S ‴ (z), {\displaystyle a(z)\cdot S(z)+b(z)\cdot zS'(z)+c(z)\cdot z^{2}S''(z)+d(z)\cdot z^{3}S'''(z),} for a (z) = 6(1 − 3 z), b (z) = 18(1 − 3 z), c (z) = 9(1 − 3 z), and d (z) = (1 − 3 z), where (1 − 3 z) = 1 − 9 z + 27 z − 27 z.

Finally, it follows that we may express the second sums through the first sums in the following form: s ~ n = (6 (1 − 3 z) 3 ∑ n = 0 ∞ s n z n + 18 (1 − 3 z) 3 ∑ n = 0 ∞ n s n z n + 9 (1 − 3 z) 3 ∑ n = 0 ∞ n (n − 1) s n z n + (1 − 3 z) 3 ∑ n = 0 ∞ n (n − 1) (n − 2) s n z n) = (n + 1) (n + 2) (n + 3) s n − 9 n (n + 1) (n + 2) s n − 1 + 27 (n − 1) n (n + 1) s n − 2 − (n − 2) (n − 1) n s n − 3. {\displaystyle {\begin{aligned}{\tilde {s}}_{n}&=\left(6(1-3z)^{3}\sum _{n=0}^{\infty }s_{n}z^{n}+18(1-3z)^{3}\sum _{n=0}^{\infty }ns_{n}z^{n}+9(1-3z)^{3}\sum _{n=0}^{\infty }n(n-1)s_{n}z^{n}+(1-3z)^{3}\sum _{n=0}^{\infty }n(n-1)(n-2)s_{n}z^{n}\right)\\&=(n+1)(n+2)(n+3)s_{n}-9n(n+1)(n+2)s_{n-1}+27(n-1)n(n+1)s_{n-2}-(n-2)(n-1)ns_{n-3}.\end{aligned}}}