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# help

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In a sequence of ten terms, each term (starting with the third term) is equal to the sum of the two previous terms. The seventh term is equal to 6. Find the sum of all ten terms.

Mar 5, 2019

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+6046
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$$\text{If you grind through it using }a_1,~a_2 \text{ as the first two sequence elements you get}\\ s =\left\{a_1,a_2,a_1+a_2,a_1+2 a_2,2 a_1+3 a_2,3 a_1+5 a_2,5 a_1+8 a_2,8 a_1+13 a_2,13 a_1+21 a_2,21 a_1+34 a_2\right\}\\ \text{Summing we get }\\ \sum \limits_{n=1}^{10}~s_k = 55 a_1+88 a_2\\ s_7=5 a_1+8 a_2 = 6\\ a_2 = \dfrac{1}{8} \left(6-5 a_1\right)\\ \text{dumping this into the sum we find (like magic!)}\\ \sum \limits_{n=1}^{10}~s_k = 66$$

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Mar 5, 2019
#2
+23835
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In a sequence of ten terms, each term (starting with the third term) is equal to the sum of the two previous terms.

The seventh term is equal to 6. Find the sum of all ten terms.

Let Fibonacci number sequence is:

$$\begin{array}{|rcll|} \hline \ldots \\ f_{-1} &=& 1 \\ f_{0} &=& 0 \\ f_{1} &=& 1 \\ f_{2} &=& 1 \\ f_{3} &=& 2 \\ f_{4} &=& 3 \\ f_{5} &=& 5 \\ f_{6} &=& 8 \\ f_{7} &=& 13 \\ f_{8} &=& 21 \\ f_{9} &=& 34 \\ f_{10} &=& 55 \\ f_{11} &=& 89 \\ f_{12} &=& 144 \\ \ldots \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline a_n &=& f_{n-2}a_1+f_{n-1}a_2 \\ s_n &=& f_na_1+ \left(f_{n+1}-1 \right)a_2\\ \hline a_{7} = f_5a_1+f_6a_2 &=& 6 \\ f_6a_2 &=& 6-f_5a_1 \\ a_2 &=& \frac{6-f_5a_1}{f_6} \\\\ s_{10} &=& f_{10}a_1+ \left(f_{11}-1\right)a_2\\ &=& f_{10}a_1+ \left(f_{11}-1\right)\left(\frac{6-f_5a_1}{f_6}\right)\\ &=& f_{10}a_1+ \dfrac{6\left(f_{11}-1\right)}{f_6}- \dfrac{f_5\left(f_{11}-1\right)}{f_6}a_1 \\ &=& \left(f_{10}- \dfrac{f_5\left(f_{11}-1\right)}{f_6}\right)a_1+ \dfrac{6\left(f_{11}-1\right)}{f_6} \\ &=& \left(55- \dfrac{5\left(89-1\right)}{8}\right)a_1+ \dfrac{6\left(89-1\right)}{8} \\ &=& \left(55- \dfrac{5\cdot 88}{8}\right)a_1+ \dfrac{6\cdot 88}{8} \\ &=& (55- 55 )a_1+ 6\cdot 11 \\ &=& 0+66 \\ \mathbf{s_{10}} &\mathbf{=}& \mathbf{ 66 } \\ \hline \end{array}$$

Mar 5, 2019