If the ordinary generating function of a sequence \(\left \{a_n\right…
2017
If the ordinary generating function of a sequence \(\left \{a_n\right \}_{n=0}^\infty\) is \(\large \frac{1+z}{(1-z)^3}\), then \(a_3-a_0\) is equal to ___________ .
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Correct answer: 15
Key idea: Use the binomial expansion for 1/(1 - z)^3.
Expand 1/(1 - z)^3 as a power series: the coefficient of z^n is C(n+2, 2), i.e. 1/(1 - z)^3 = Σ_{n≥0} C(n+2,2) z^n.
Multiply by (1 + z). The coefficient a_n of z^n in (1+z)/(1-z)^3 is C(n+2,2) + C(n+1,2).
Compute a_0: C(2,2) + C(1,2) = 1 + 0 = 1.
Compute a_3: C(5,2) + C(4,2) = 10 + 6 = 16.
Therefore a_3 - a_0 = 16 - 1 = 15.
Answer: 15