Choose one of the following to write up. (You should be able to prove any of the
ID: 3142694 • Letter: C
Question
Choose one of the following to write up. (You should be able to prove any of them): (a) Let a_n be the sequence defined by a_1 = 1, a_2 = 8, and a_n = a_n-1 + 2a_n - 2 for n greaterthanorequalto 3. i. Write down the first 5 terms of the sequence ii. Prove that for all n elementof N, a_n = 3 middot 2^n - 1 + 2(-l)^n. (b) Let b_n be the sequence defined by b_1 = 1, b_2 = 2, b_3 = 3, and b_n = b_n-1 + b_n - 2 + b_n - 3 for n greaterthanorequalto 4. This sequence is sometimes referred to as a Tribonacci sequence, with signature (1, 2, 3). Show that b_n lessthanorequalto 3^n. (c) Concerning the Fibonacci sequence, prove that F_2 + F_4 + F_6 + .. + F_2n = F_2n + 1 - 1.Explanation / Answer
(C) Fibonacci series= 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, . . .
Each Fibonacci number is the sum of the previous two Fibonacci numbers.
For any positive integer n, the Fibonacci numbers satisfy: F1 + F2 + F3 + · · · + Fn = Fn+2 1
Let's call even Fibonaccis, since their index is EVEN.
F2=1, F4=3, F6=8, F20=?
nth even Fibonacci number= F2n
let's take some few starting numbers of series,
F2 + F4 = 1 + 3 = 4 = F5 1
F2 + F4 + F6 = 1 + 3 + 8 = 12 = F7 1
F2 + F4 + F6 + F8 = 1 + 3 + 8 + 21 = 33 = F9 1
F2 + F4 + F6 + F8 + F10 = 1 + 3 + 8 + 21 + 55 = 88 = F11 1
It looks like the sum of the first few “even” Fibonacci numbers is one less than another Fibonacci number.
The sum of the first 5 even Fibonacci numbers (up to F10) is the 11th Fibonacci number less one.
Maybe it’s true that the sum of the first n “even” Fibonacci’s is one less than the next Fibonacci number. That is, For any positive integer n, the Fibonacci numbers satisfy:
F2 + F4 + F6 + · · · + F2n = F2n+1 1
Proof:
we know,
F5 = F4 + F3.
So: F4 = F5 F3
In general: Feven = Fnext odd Fprevious odd
F2 = F3 F1
F4 = F5 F3
F6 = F7 F5
F8 = F9 F7 . . . . . . F2n2 = F2n1 F2n3 F2n = F2n+1 F2n1
F2 = F3 F1
F4 = F5 F3
F6 = F7 F5
F8 = F9 F7 . . . . . .
F2n2 = F2n1 F2n3 + F2n = F2n+1 F2n1
after adding above terms, F3, F5, F7, F9, F2n-1, F2n-1 get cancalled.
and we get,
F2 + F4 + F6 + · · · + F2n = F2n+1 F1
F1=1, by substituting, we get
F2 + F4 + F6 + · · · + F2n = F2n+1 1
Hence proved.