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# Help

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182
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+133

Let $$f_1,f_2,f_3,...$$ be a sequence of numbers such that $$f_n = f_{n - 1} + f_{n - 2}$$
for every integer $$n \ge 3$$ . If $$f_7 = 83$$, what is the sum of the first 10 terms of the sequence?

Jun 18, 2019

#1
+1

Here is one possibility. there maybe others:
5, 7.25, 12.25, 19.5, 31.75, 51.25, 83, 134.25, 217.25, 351.50 =913

Jun 18, 2019
#2
+106519
+2

We can set this up as a Fibonacci Series

f1 =f1

f2 = f2

f3 = f2 + f1

f4 = 2f2 + f1

f5 = 3f2 + 2f1

f6 = 5f2 + 3f1

f7 = 8f2 + 5f1

f8 = 13f2 + 8f1

f9 = 21f2 + 13f1

f10 = 34f2 + 21fi

______________

Sum  =  88f2 + 55f1

Using a Fibonacci identity......  11f(5)  =  f(10)

So

11 [ 3f2 + 2f1]  =  33f2 + 22f1   =  34f2 + 21f1 ⇒   f1 = f2

Then

f7  =  8f2 + 5f1

f7 = 8f1 + 5fi

f7 = 13f1  = 83

13f1  = 83

f1 = 83/13

And the sum of the series =   88f2 + 55f1  =  88f1 + 55f1  =   143 f1

So....the sum is    143 (83/13)  =   913

Just as the Guest found  !!!

Jun 18, 2019
edited by CPhill  Jun 18, 2019
#3
+133
+1

Thanks

Jun 19, 2019
#4
+23805
+1

Let $$f_1,f_2,f_3,\ldots$$ be a sequence of numbers such that $$f_n = f_{n - 1} + f_{n - 2}$$ for every integer $$n \ge 3$$ .

If $$f_7 = 83$$, what is the sum of the first 10 terms of the sequence?

$$\begin{array}{|rclcl|} \hline f_1 &=& 0\cdot f_2 &+& 1\cdot f_1 \\ f_2 &=& 1\cdot f_2 &+& 0\cdot f_1 \\ f_3 &=& 1\cdot f_2 &+& 1\cdot f_1 \\ f_4 &=& 2\cdot f_2 &+& 1\cdot f_1 \\ f_5 &=& 3\cdot f_2 &+& 2\cdot f_1 \\ f_6 &=& 5\cdot f_2 &+& 3\cdot f_1 \\ f_7 &=& 8\cdot f_2 &+& 5\cdot f_1 \\ \ldots \\ \mathbf{f_n} &=& \mathbf{ F_{n-1} \cdot f_2} &\mathbf{+}& \mathbf{F_{n-2}\cdot f_1 } \\\\ && && \text{Fibonacci Numbers:} \\ && && \ldots \\ && && F_{-1} = 1 \\ && && F_{0} = 0 \\ && && F_{1} = 1 \\ && && F_{2} = 1 \\ && && F_{3} = 2 \\ && && F_{4} = 3 \\ && && F_{5} = 5 \\ && && F_{6} = 8 \\ && && F_{7} = 13 \\ && && F_{8} =21 \\ && && F_{9} = 34 \\ && && F_{10} = 55 \\ && && F_{11} = 89 \\ && && \ldots \\ \hline \end{array}$$

$$\mathbf{f_1 = \ ?,\ f_2 = \ ?}$$

$$\begin{array}{|rcll|} \hline f_7 = 83 &=& F_6\cdot f_2 + F_5\cdot f_1 \quad | \quad F_6 = 8,\qquad F_5=5 \\ \hline \end{array}$$

$$\begin{array}{|rclrclrclrcl|} \hline 83 &=& 8f_2 + 5f_1 \\ 5f_1 &=&83- 8f_2 \\ f_1 &=& \dfrac{83- 8f_2}{5} \\ f_1 &=& \dfrac{80-5f_2-3f_2+3}{5} \\ f_1 &=& 16-f_2+\underbrace{\dfrac{3-3f_2}{5}}_{=a} \\ f_1 &=& 16-f_2+a & a &=& \dfrac{3-3f_2}{5} \\ & & & 5a &=& 3-3f_2 \\ & & & 3f_2 &=& 3-5a \\ & & & f_2 &=& \dfrac{3-5a}{3} \\ & & & f_2 &=& \dfrac{3-3a-2a}{3} \\ & & & f_2 &=& 1-a-\underbrace{\dfrac{2a}{3}}_{=b} \\ & & & f_2 &=& 1-a-b & b &=& \dfrac{2a}{3} \\ & & & & & & 3b &=& 2a \\ & & & & & & 2a &=& 3b \\ & & & & & & a &=& \dfrac{3b}{2} \\ & & & & & & a &=& \dfrac{2b+b}{2} \\ & & & & & & a &=& b +\underbrace{\dfrac{b}{2}}_{=c} \\ & & & & & & a &=& b +c & c &=& \dfrac{b}{2} \\ & & & & & & & & & \mathbf{b} &=& \mathbf{2c} \\ & & & & & & a &=& 2c +c \\ & & & & & & \mathbf{a} &=& \mathbf{3c} \\ & & & f_2 &=& 1-3c-2c \\ & & & \mathbf{f_2} &=& \mathbf{1-5c} \\ f_1 &=& 16-(1-5c)+3c \\ \mathbf{f_1} &=& \mathbf{15+8c} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{f_1} &=& \mathbf{15+8c} ,\ c\in \mathbb{Z}\\ \mathbf{f_2} &=& \mathbf{1-5c},\ c\in \mathbb{Z}\\ \hline \end{array}$$

sum:

$$\begin{array}{|rclcl|} \hline s_1 &=& 0\cdot f_2 &+& 1\cdot f_1 \\ s_2 &=& 1\cdot f_2 &+& 1\cdot f_1 \\ s_3 &=& 2\cdot f_2 &+& 2\cdot f_1 \\ s_4 &=& 4\cdot f_2 &+& 3\cdot f_1 \\ s_5 &=& 7\cdot f_2 &+& 5\cdot f_1 \\ s_6 &=& 20\cdot f_2 &+& 8\cdot f_1 \\ \ldots \\ \mathbf{s_n} &=& \mathbf{ (F_{n+1}-1) \cdot f_2} &\mathbf{+}& \mathbf{F_{n}\cdot f_1 } \\ \hline \end{array}$$

$$\mathbf{s_{10}=\ ?}$$

$$\begin{array}{|rcll|} \hline \mathbf{s_{10}} &=& \mathbf{(F_{11}-1)\cdot f_2 +F_{10}} \cdot f_1 \quad &| \quad F_{11}=89,\ F_{10}= 55 \\\\ &=& 88 f_2 +55f_1 \quad &| \quad f_1 = 15+8c,\ f_2 = 1-5c \\ &=& 88\cdot(1-5c) +55\cdot(15+8c) \\ &=& 88 -88\cdot 5c +55\cdot 15+ 55\cdot 8c \quad &| \quad 88\cdot 5 = 55\cdot 8 \\ &=& 88 +55\cdot 15 \\ &=& \mathbf{913} \\ \hline \end{array}$$

The sum of the first 10 terms of the sequence is 913

Jun 19, 2019