+1124 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!!..
+26393 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?
\(\begin{array}{|rcll|} \hline && \text{Let } \mathcal{F} \text{ are the Fibonacci numbers }\\\\ \mathcal{F}_1 &=& 1\\ \mathcal{F}_2 &=& 1\\ \mathcal{F}_3 &=& 2\\ \mathcal{F}_4 &=& 3\\ \mathcal{F}_5 &=& 4\\ \mathcal{F}_6 &=& 5\\ \mathcal{F}_7 &=& 13\\ \mathcal{F}_8 &=& 21\\ \mathcal{F}_9 &=& 34\\ \mathcal{F}_{10} &=& 55\\ \mathcal{F}_{11} &=& 89\\ \mathcal{F}_{12} &=& 144\\ \ldots \\ \hline \end{array} \)
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
+26393 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?
\(\begin{array}{|rcll|} \hline && \text{Let } \mathcal{F} \text{ are the Fibonacci numbers }\\\\ \mathcal{F}_1 &=& 1\\ \mathcal{F}_2 &=& 1\\ \mathcal{F}_3 &=& 2\\ \mathcal{F}_4 &=& 3\\ \mathcal{F}_5 &=& 4\\ \mathcal{F}_6 &=& 5\\ \mathcal{F}_7 &=& 13\\ \mathcal{F}_8 &=& 21\\ \mathcal{F}_9 &=& 34\\ \mathcal{F}_{10} &=& 55\\ \mathcal{F}_{11} &=& 89\\ \mathcal{F}_{12} &=& 144\\ \ldots \\ \hline \end{array} \)
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
+1124 Good heavens Heureka,
this is WWAAAAAYYYYY over my comprehension!!!..thank you soo much!!
