Using induction on the positive integer n, establish the following formulas: (a) #+2u2+3u3 + +nu, = (n + 1)un+2 - Un+4 +2. (6) 2+2u4 +3u6 + (a) Show thot | ...+ nuzn = nu2n+1-U2n. %3D

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Chapter2: Second-order Linear Odes
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14.3 question 1

4 (mod 2p- 1)
5-1=5@-22 (5/2p-1) (mod 2p 1)
Eom Theorems 9.9 and 9.10, it is easy to see that
(5/2p-1)%3D(2p-1/5)3D (10k +3/5) = (3/5)%3D-1
ve arive at 2u = 1+(-1)= 0 (mod 2p-1), from which it may be conclu
sup. The case p = 4 (mod 5) can be handled in much the same .
that 2p - 1 divides
upon noting that (2/5) =-1.
iustrations, we mention u 19 = 37 113, where 19 = 4 (mod 5); and u-
73-330929, where 37 2 (mod 5).
The Fibonacci numbers provide a continuing source of questions for invest
- Here is a recent result: the largest Fibonacci number that is the sum of two
ials is u12= 144=4! +5!. Another is that the only squares among the Fibor
mbers are u = 1 and u12 = 12", with the only other power being ug =
= 23.
%3D
%3D
PROBLEMS 14.3
L. Using induction on the positive integer n, establish the following formulas:
(a) #1+2u2+3uz + +nun = (n+ 1)un+2 – Un+4 + 2.
%3D
2(@) Show that the sum of the first n Fibonacci numbers with odd indices is given
formula
%3D
uzn - I+7nu = "Tnu +….+9nɛ + + ?n (a)
%3D
zn =110+ + Sn+ En + In
%3D
LAint: Add the equalities uj = u2, uz = U4 - u2, u5 = U6- 44,· ·
Stow that the sum of the first n Fibonacci numbers with even indices is given
formula
%3D
%3D
%3D
L- I+4zn = n +..+ In +4n+n
Transcribed Image Text:4 (mod 2p- 1) 5-1=5@-22 (5/2p-1) (mod 2p 1) Eom Theorems 9.9 and 9.10, it is easy to see that (5/2p-1)%3D(2p-1/5)3D (10k +3/5) = (3/5)%3D-1 ve arive at 2u = 1+(-1)= 0 (mod 2p-1), from which it may be conclu sup. The case p = 4 (mod 5) can be handled in much the same . that 2p - 1 divides upon noting that (2/5) =-1. iustrations, we mention u 19 = 37 113, where 19 = 4 (mod 5); and u- 73-330929, where 37 2 (mod 5). The Fibonacci numbers provide a continuing source of questions for invest - Here is a recent result: the largest Fibonacci number that is the sum of two ials is u12= 144=4! +5!. Another is that the only squares among the Fibor mbers are u = 1 and u12 = 12", with the only other power being ug = = 23. %3D %3D PROBLEMS 14.3 L. Using induction on the positive integer n, establish the following formulas: (a) #1+2u2+3uz + +nun = (n+ 1)un+2 – Un+4 + 2. %3D 2(@) Show that the sum of the first n Fibonacci numbers with odd indices is given formula %3D uzn - I+7nu = "Tnu +….+9nɛ + + ?n (a) %3D zn =110+ + Sn+ En + In %3D LAint: Add the equalities uj = u2, uz = U4 - u2, u5 = U6- 44,· · Stow that the sum of the first n Fibonacci numbers with even indices is given formula %3D %3D %3D L- I+4zn = n +..+ In +4n+n
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