Which of the following is the closed form of the ordinary generating function…
2022
Which of the following is the closed form of the ordinary generating function
\(G(x)=\sum_{n=0}^{\infty} a_nx^n\) for the sequence \(\{a_n\}_{n\geq 0}\)
defined below?
\(a_n=n+1\), if \(n\) is odd; otherwise, \(a_n=1\).
Answer: A. \(\frac {x(1 + x^2)} {(1 - x^2)^2} + \frac {1} {1 - x}\) — Key idea: separate the constant 1 from the odd-index contributions. Write the sequence as a(n)=1 for every n, plus an extra n when n is odd. Thus…
- A.
\(\frac {x(1 + x^2)} {(1 - x^2)^2} + \frac {1} {1 - x}\) - B.
\(\frac {x(3 - x^2)} {(1 - x^2)^2} + \frac {1} {1 - x}\) - C.
\(\frac {2x} {(1 - x^2)^2} + \frac {1} {1 - x}\) - D.
\(\frac {x} {(1 - x^2)^2} + \frac {1} {1 - x}\)
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Correct answer: A
Key idea: separate the constant 1 from the odd-index contributions.
Write the sequence as a(n)=1 for every n, plus an extra n when n is odd. Thus
G(x)=sum_{n>=0} a(n)x^n = sum_{n>=0}1*x^n + sum_{odd n} n x^n = 1/(1-x) + sum_{k>=0} (2k+1) x^{2k+1}.
Evaluate the odd-index sum: sum_{k>=0}(2k+1)x^{2k+1} = x*(2*sum_{k>=0} k (x^2)^k + sum_{k>=0} (x^2)^k).
Use standard sums: sum_{k>=0} (x^2)^k = 1/(1-x^2) and sum_{k>=0} k (x^2)^k = x^2/(1-x^2)^2, so the odd sum equals x*(2*x^2/(1-x^2)^2 + 1/(1-x^2)) = x(1+x^2)/(1-x^2)^2.
Therefore the generating function is
G(x) = x(1+x^2)/(1-x^2)^2 + 1/(1-x).
Quick check: expanding the series gives coefficients 1,2,1,4,1,6,... which match a0,a1,a2,a3,a4,a5 respectively.
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