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Theorem 2.4. If pn and qn are defined by po = ao, p₁ = a1a0 +1, Pn = anpn−1 + Pn-2 for 2≤ n 901, 91a1, In = angn−1 + In-2 for 2≤ n, = then Pn [ao, a1,..., an] = In Theorem 2.5. The numbers pn and qn satisfy Pngn-1-Pn-19n = (-1) "-1 Proof. From the previous theorem, we have PnIn-1 Pn-19n = (anPn-1 + Pn-2)In-1 - Pn-1(anIn−1 + In-2) =-(Pn-19n-2-Pn-29n-1) Repeating this step with n-1, n - 2, ..., 2 in place of n, gives us 9 PnIn-1 — Pn-19n = (-1)-1 (P190 - Po91) = (-1)"-1(1) = (-1)"-1 Q.E.D.

Theorem 2.4. If pn and qn are defined by po = ao, p₁ = a1a0 +1, Pn = anpn−1 + Pn-2 for 2≤ n 901, 91a1, In = angn−1 + In-2 for 2≤ n, = then Pn [ao, a1,..., an] = In Theorem 2.5. The numbers pn and qn satisfy Pngn-1-Pn-19n = (-1) "-1 Proof. From the previous theorem, we have PnIn-1 Pn-19n = (anPn-1 + Pn-2)In-1 - Pn-1(anIn−1 + In-2) =-(Pn-19n-2-Pn-29n-1) Repeating this step with n-1, n - 2, ..., 2 in place of n, gives us 9 PnIn-1 — Pn-19n = (-1)-1 (P190 - Po91) = (-1)"-1(1) = (-1)"-1 Q.E.D.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.5: The Binomial Theorem
Problem 16E
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Please fill in the rest of the steps of the proof of Thm 2.5. Show how "Repeating this step with n-1,n-2,...,2 in place of n" gives us the desired result.

Theorem 2.4. If pn and qn are defined by po = ao, p₁ = a1a0 +1, Pn = anpn−1 + Pn-2 for 2≤ n 901, 91a1, In = angn−1 + In-2 for 2≤ n, = then Pn [ao, a1,..., an] = In
Transcribed Image Text:Theorem 2.4. If pn and qn are defined by po = ao, p₁ = a1a0 +1, Pn = anpn−1 + Pn-2 for 2≤ n 901, 91a1, In = angn−1 + In-2 for 2≤ n, = then Pn [ao, a1,..., an] = In
Theorem 2.5. The numbers pn and qn satisfy Pngn-1-Pn-19n = (-1) "-1 Proof. From the previous theorem, we have PnIn-1 Pn-19n = (anPn-1 + Pn-2)In-1 - Pn-1(anIn−1 + In-2) =-(Pn-19n-2-Pn-29n-1) Repeating this step with n-1, n - 2, ..., 2 in place of n, gives us 9 PnIn-1 — Pn-19n = (-1)-1 (P190 - Po91) = (-1)"-1(1) = (-1)"-1 Q.E.D.
Transcribed Image Text:Theorem 2.5. The numbers pn and qn satisfy Pngn-1-Pn-19n = (-1) "-1 Proof. From the previous theorem, we have PnIn-1 Pn-19n = (anPn-1 + Pn-2)In-1 - Pn-1(anIn−1 + In-2) =-(Pn-19n-2-Pn-29n-1) Repeating this step with n-1, n - 2, ..., 2 in place of n, gives us 9 PnIn-1 — Pn-19n = (-1)-1 (P190 - Po91) = (-1)"-1(1) = (-1)"-1 Q.E.D.
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