+0

# Hi guys,

0
195
3
+313

Hi guys,

i have just one more question please....

A sequence of 2016 terms is formed as follows: The first two terms are equal to 3, from the 3rd term onwards, each consecutive term is formed by finding the sum of the previous two terms. If each of the terms of the sequence 3,3,6,9....are now devided by 2, and the remainders are added, what will the sum of the first 2016 remainders be?

all and any help will be greatly appreciated!!..

Jul 25, 2018

#1
+21242
+2

A sequence of 2016 terms is formed as follows: The first two terms are equal to 3, from the 3rd term onwards, each consecutive term is formed by finding the sum of the previous two terms.

If each of the terms of the sequence 3,3,6,9....are now devided by 2,

what will the sum of the first 2016 remainders be?

$$\begin{array}{|rcll|} \hline && \text{Let } \mathcal{F} \text{ are the Fibonacci numbers }\\\\ \mathcal{F}_1 &=& 1\\ \mathcal{F}_2 &=& 1\\ \mathcal{F}_3 &=& 2\\ \mathcal{F}_4 &=& 3\\ \mathcal{F}_5 &=& 4\\ \mathcal{F}_6 &=& 5\\ \mathcal{F}_7 &=& 13\\ \mathcal{F}_8 &=& 21\\ \mathcal{F}_9 &=& 34\\ \mathcal{F}_{10} &=& 55\\ \mathcal{F}_{11} &=& 89\\ \mathcal{F}_{12} &=& 144\\ \ldots \\ \hline \end{array}$$

The sequence:

$$\begin{array}{|rccr|l|} \hline a_1 &=& &=& 1\cdot 3 \\ a_2 &=& &=& 1\cdot 3 \\ a_3 &=& 1\cdot 3 + 1\cdot 3 &=& 2\cdot 3 \\ a_4 &=& 1\cdot 3 + 2\cdot 3 &=& 3\cdot 3 \\ a_5 &=& 2\cdot 3 + 3\cdot 3 &=& 5\cdot 3 \\ a_6 &=& 3\cdot 3 + 5\cdot 3 &=& 8\cdot 3 \\ a_7 &=& 5\cdot 3 + 8\cdot 3 &=& 13\cdot 3 \\ a_8 &=& 8\cdot 3 + 13\cdot 3 &=& 21\cdot 3 \\ a_9 &=& 13\cdot 3 + 21\cdot 3 &=& 34\cdot 3 \\ \ldots \\ a_n &=& && \mathcal{F}_{n} \cdot 3 \\ \hline \end{array}$$

$$\begin{array}{|r|rcl|c|} \hline \text{cycle} && && \text{remainder after divided by } 2 \\ \hline 1 & a_1 &=& 1\cdot 3 & 1 \\ 1 & a_2 &=& 1\cdot 3 & 1 \\ 1 & a_3 &=& 2\cdot 3 & 0 \\ \hline 2 & a_4 &=& 3\cdot 3 & 1 \\ 2 & a_5 &=& 5\cdot 3 & 1 \\ 2 & a_6 &=& 8\cdot 3 & 0 \\ \hline 3 & a_7 &=& 13\cdot 3 & 1 \\ 3 & a_8 &=& 21\cdot 3 & 1 \\ 3 & a_9 &=& 34\cdot 3 & 0 \\ \hline \ldots \\ \hline 672 & a_{2014} &=& \mathcal{F}_{2014} \cdot 3 & 1 \\ 672 & a_{2015} &=& \mathcal{F}_{2015} \cdot 3 & 1 \\ 672 & a_{2016} &=& \mathcal{F}_{2016} \cdot 3 & 0 \\ \hline &&&& \text{sum }=672*2 = 1344 \\ \hline \end{array}$$

The sum of the first 2016 remainders will be 1344

Jul 25, 2018
edited by heureka  Jul 25, 2018

#1
+21242
+2

A sequence of 2016 terms is formed as follows: The first two terms are equal to 3, from the 3rd term onwards, each consecutive term is formed by finding the sum of the previous two terms.

If each of the terms of the sequence 3,3,6,9....are now devided by 2,

what will the sum of the first 2016 remainders be?

$$\begin{array}{|rcll|} \hline && \text{Let } \mathcal{F} \text{ are the Fibonacci numbers }\\\\ \mathcal{F}_1 &=& 1\\ \mathcal{F}_2 &=& 1\\ \mathcal{F}_3 &=& 2\\ \mathcal{F}_4 &=& 3\\ \mathcal{F}_5 &=& 4\\ \mathcal{F}_6 &=& 5\\ \mathcal{F}_7 &=& 13\\ \mathcal{F}_8 &=& 21\\ \mathcal{F}_9 &=& 34\\ \mathcal{F}_{10} &=& 55\\ \mathcal{F}_{11} &=& 89\\ \mathcal{F}_{12} &=& 144\\ \ldots \\ \hline \end{array}$$

The sequence:

$$\begin{array}{|rccr|l|} \hline a_1 &=& &=& 1\cdot 3 \\ a_2 &=& &=& 1\cdot 3 \\ a_3 &=& 1\cdot 3 + 1\cdot 3 &=& 2\cdot 3 \\ a_4 &=& 1\cdot 3 + 2\cdot 3 &=& 3\cdot 3 \\ a_5 &=& 2\cdot 3 + 3\cdot 3 &=& 5\cdot 3 \\ a_6 &=& 3\cdot 3 + 5\cdot 3 &=& 8\cdot 3 \\ a_7 &=& 5\cdot 3 + 8\cdot 3 &=& 13\cdot 3 \\ a_8 &=& 8\cdot 3 + 13\cdot 3 &=& 21\cdot 3 \\ a_9 &=& 13\cdot 3 + 21\cdot 3 &=& 34\cdot 3 \\ \ldots \\ a_n &=& && \mathcal{F}_{n} \cdot 3 \\ \hline \end{array}$$

$$\begin{array}{|r|rcl|c|} \hline \text{cycle} && && \text{remainder after divided by } 2 \\ \hline 1 & a_1 &=& 1\cdot 3 & 1 \\ 1 & a_2 &=& 1\cdot 3 & 1 \\ 1 & a_3 &=& 2\cdot 3 & 0 \\ \hline 2 & a_4 &=& 3\cdot 3 & 1 \\ 2 & a_5 &=& 5\cdot 3 & 1 \\ 2 & a_6 &=& 8\cdot 3 & 0 \\ \hline 3 & a_7 &=& 13\cdot 3 & 1 \\ 3 & a_8 &=& 21\cdot 3 & 1 \\ 3 & a_9 &=& 34\cdot 3 & 0 \\ \hline \ldots \\ \hline 672 & a_{2014} &=& \mathcal{F}_{2014} \cdot 3 & 1 \\ 672 & a_{2015} &=& \mathcal{F}_{2015} \cdot 3 & 1 \\ 672 & a_{2016} &=& \mathcal{F}_{2016} \cdot 3 & 0 \\ \hline &&&& \text{sum }=672*2 = 1344 \\ \hline \end{array}$$

The sum of the first 2016 remainders will be 1344

heureka Jul 25, 2018
edited by heureka  Jul 25, 2018
#2
+313
+1

Good heavens Heureka,

this is WWAAAAAYYYYY over my comprehension!!!..thank you soo much!!

juriemagic  Jul 25, 2018
#3
+1

If you have difficulty understanding what heureka did, just use this summation of the sequence which will give you the same answer:

∑[ 3 F_n mod 2, n, 1, 2016] =1,344, where F_n =the nth Fibonacci number.

Jul 25, 2018