duce the expression for u to 2(5 ,Y = (1+5P) +4 (mod 2p - 1) or simply, to 2u, = 1+5P-1(mod 2p - 1). Now 5P-1 =52p-2)/2 = (5/2p- 1)(mod 2p- 1) %3D From Theorems 9.9 and 9.10, it is easy to see that (5/2p- 1) = (2p - 1/5) = (10k + 3/5) = (3/5) = -1 %3D Last, we arrive at 2u = 1+(-1) = 0(mod 2p -1 divides up. The casep = 4 (mod 5) can be handled in much the same way -1), from which it may be concluded that 2p upon noting that (2/5) = -1. As illustrations, we mention u 19 = 37 · 113, where 19 = 4 (mod 5); and u37 = 73-330929, where 37 = 2 (mod 5). The Fibonacci numbers provide a continuing source of questions for investiga- tion. Here is a recent result: the largest Fibonacci number that is the sum of two fac- torials is u 12 numbers are u =1 and u12 = 12', with the only other power being u6 = 2³. 3D144%3D4!+5!. Another is that the only squares among the Fibonacci PROBLEMS 14.3 1. Using induction on the positive integer n, establish the following formulas: (a) u +2u2 +3u3 + +nun = (n + 1)u,n+2- Un+4 + 2. (b) u2 + 2u4 + 3u6 +...+ nu2n = nun+1 – U2n 2. (a) Show that the sum of the first n Fibonacci numbers with odd indices is given by the formula 7n = +….+ Sn+ €n + In 0) Show that the sum of the first n Fibonacci numbers with even indices is given by the formula Hint: Add the equalities uj = u2, U3 = U4 – u2, U5 = U6 –U4, THìnt: Apply part (a) in conjunction with identity in Eq. (2).J
duce the expression for u to 2(5 ,Y = (1+5P) +4 (mod 2p - 1) or simply, to 2u, = 1+5P-1(mod 2p - 1). Now 5P-1 =52p-2)/2 = (5/2p- 1)(mod 2p- 1) %3D From Theorems 9.9 and 9.10, it is easy to see that (5/2p- 1) = (2p - 1/5) = (10k + 3/5) = (3/5) = -1 %3D Last, we arrive at 2u = 1+(-1) = 0(mod 2p -1 divides up. The casep = 4 (mod 5) can be handled in much the same way -1), from which it may be concluded that 2p upon noting that (2/5) = -1. As illustrations, we mention u 19 = 37 · 113, where 19 = 4 (mod 5); and u37 = 73-330929, where 37 = 2 (mod 5). The Fibonacci numbers provide a continuing source of questions for investiga- tion. Here is a recent result: the largest Fibonacci number that is the sum of two fac- torials is u 12 numbers are u =1 and u12 = 12', with the only other power being u6 = 2³. 3D144%3D4!+5!. Another is that the only squares among the Fibonacci PROBLEMS 14.3 1. Using induction on the positive integer n, establish the following formulas: (a) u +2u2 +3u3 + +nun = (n + 1)u,n+2- Un+4 + 2. (b) u2 + 2u4 + 3u6 +...+ nu2n = nun+1 – U2n 2. (a) Show that the sum of the first n Fibonacci numbers with odd indices is given by the formula 7n = +….+ Sn+ €n + In 0) Show that the sum of the first n Fibonacci numbers with even indices is given by the formula Hint: Add the equalities uj = u2, U3 = U4 – u2, U5 = U6 –U4, THìnt: Apply part (a) in conjunction with identity in Eq. (2).J
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
14.3 question 2
Transcribed Image Text:is
duce the expression for u, to
2(5up) 3D(1+ 5P) +4 (mod 2p- 1)
or simply, to 2u, =1+5P-(mod 2p - 1). Now
From Theorems 9.9 and 9.10, it is easy to see that
(5/2p-1)%3D(2p - 1/5) = (10k +3/5) = (3/5) = -1
Lest we arrive at 2u = 1+(-1)30 (mod 2p - 1), from which it may be concluded
that 2p -1 divides u p. The case p = 4 (mod 5) can be handled in much the same way
upon noting that (2/5) = -1.
%3D
=
As illustrations, we mention u19 =
73-330929, where 37 = 2 (mod 5).
The Fibonacci numbers provide a continuing source of questions for investiga-
tion Here is a recent result: the largest Fibonacci number that is the sum of two fac-
torials is u 12 = 144 = 4! + 5!. Another is that the only squares among the Fibonacci
37 · 113, where 19 = 4 (mod 5); and u37 =
%3D
%3D
122, with the only other power being u6 = 2.
numbers are u 1=1 and u12
PROBLEMS 14.3
1. Using induction on the positive integer n, establish the following formulas:
(a) u+2u2 + 3u3 +. +nun = (n + 1)un+2 - Un+4 + 2.
(b) u2+2u4+ 3u6 +
2. (a) Show that the sum of the first n Fibonacci numbers with odd indices is given by the
formula
%3D
+ nu2n = nu2n+1
Ut+3 + u5+ +u2n-1 = U2n
u2, U3 = U4 – u2, U5 = U6 – 44,-
show that the sum of the first n Fibonacci numbers with even indices is given by the
formula
%3D
[Hint: Add the equalities u1 =
%3D
Hint: Apply part (a) in conjunction with identity in Eq. (2).J
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