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The sequence $x_1$, $x_2$, $x_3$, . . ., has the property that $x_n = x_{n - 1} + x_{n - 2}$ for all $n \ge 3$. If $x_{11} - x_1 = 99$, then determine $x_6.$

michaelcai  Oct 12, 2017

Best Answer 

 #2
avatar+19482 
+1

The sequence x_1, x_2, x_3, . . ., has the property that x_n = x_{n - 1} + x_{n - 2}

for all n \ge 3.

If x_{11} - x_1 = 99, then determine x_6.

 

\(\begin{array}{|rcll|} \hline x_3 &=& x_2 + x_1 &=& 1\cdot x_2 + 1\cdot x_1 \\ x_4 &=& x_3 + x_2 &=& 2\cdot x_2 + 1\cdot x_1 \\ x_5 &=& x_4 + x_3 &=& 3\cdot x_2 + 2\cdot x_1 \\ \color{red}x_6 &\color{red}=& x_5 + x_4 &=& \color{red}5\cdot x_2 + 3\cdot x_1 \\ x_7 &=& x_6 + x_5 &=& 8\cdot x_2 + 5\cdot x_1 \\ x_8 &=& x_7 + x_6 &=& 13\cdot x_2 + 8\cdot x_1 \\ x_9 &=& x_8 + x_7 &=& 21\cdot x_2 + 13\cdot x_1 \\ x_{10} &=& x_9 + x_8 &=& 34\cdot x_2 + 21\cdot x_1 \\\\ x_{11} &=& x_{10} + x_9 &=& 55\cdot x_2 + 34\cdot x_1 \quad & | \quad x_{11} = 99 + x_1 \\ 99 + x_1 &=& 55\cdot x_2 + 34\cdot x_1 && \quad & | \quad - x_1 \\ 99 &=& 55\cdot x_2 + 33\cdot x_1 && \quad & | \quad : 11 \\ \color{red}9 &\color{red}=& \color{red}5\cdot x_2 + 3\cdot x_1 \\\\ \hline \color{red}9 &\color{red}=& \color{red}5\cdot x_2 + 3\cdot x_1 && \quad & | \quad \color{red}x_6 = 5\cdot x_2 + 3\cdot x_1\\ \mathbf{9} &\mathbf{=}& \mathbf{x_6} \\ \hline \end{array} \)

 

laugh

heureka  Oct 12, 2017
edited by heureka  Oct 12, 2017
 #1
avatar+86869 
+1

\($x_n = x_{n - 1} + x_{n - 2}$ for all $n \ge 3$\)

 

x11  =  99 + x1

x9 + x10  =  99 + (x3 - x2)

(x7 + x8) + (x8 + x9)  =  99 + (x5 - x4) - (x3 - x1)

(x5 + x6) + (x6 + x7) + (x6 + x7) + (x7 +  x8)  =  99 + x5 - x4 - x3 + x1

3x6 + 3x7 + x8  =  99  - x4 - x3 + x1

3x6  +  3(x5 + x6)  + (x6 + x7)  = 99 - (x6 - x5) - ( x5 - x4) + (x3 - x2)

7x + 3x5 + x= 99  -x6 +x4 + x3 - x2

8x6 + 3(x7 - x6) + x7  = 99  + x4 +( x5 - x4) - (x4 - x3)

5x6 + 4x7  =  99 + x5 - x4 + x3

5x6 + 4(x5 + x6)  = 99 + x5 - (x6 - x5) + (x5 - x4)

10x6 + 4x=  99 + 3x5 -x4

10x6 + x5 = 99 - x4

10x6 + x5 + x = 99

10x6  + x =  99

11x6  = 99             divide both sides by 11

x= 9

 

 

cool cool cool

CPhill  Oct 12, 2017
edited by CPhill  Oct 12, 2017
 #2
avatar+19482 
+1
Best Answer

The sequence x_1, x_2, x_3, . . ., has the property that x_n = x_{n - 1} + x_{n - 2}

for all n \ge 3.

If x_{11} - x_1 = 99, then determine x_6.

 

\(\begin{array}{|rcll|} \hline x_3 &=& x_2 + x_1 &=& 1\cdot x_2 + 1\cdot x_1 \\ x_4 &=& x_3 + x_2 &=& 2\cdot x_2 + 1\cdot x_1 \\ x_5 &=& x_4 + x_3 &=& 3\cdot x_2 + 2\cdot x_1 \\ \color{red}x_6 &\color{red}=& x_5 + x_4 &=& \color{red}5\cdot x_2 + 3\cdot x_1 \\ x_7 &=& x_6 + x_5 &=& 8\cdot x_2 + 5\cdot x_1 \\ x_8 &=& x_7 + x_6 &=& 13\cdot x_2 + 8\cdot x_1 \\ x_9 &=& x_8 + x_7 &=& 21\cdot x_2 + 13\cdot x_1 \\ x_{10} &=& x_9 + x_8 &=& 34\cdot x_2 + 21\cdot x_1 \\\\ x_{11} &=& x_{10} + x_9 &=& 55\cdot x_2 + 34\cdot x_1 \quad & | \quad x_{11} = 99 + x_1 \\ 99 + x_1 &=& 55\cdot x_2 + 34\cdot x_1 && \quad & | \quad - x_1 \\ 99 &=& 55\cdot x_2 + 33\cdot x_1 && \quad & | \quad : 11 \\ \color{red}9 &\color{red}=& \color{red}5\cdot x_2 + 3\cdot x_1 \\\\ \hline \color{red}9 &\color{red}=& \color{red}5\cdot x_2 + 3\cdot x_1 && \quad & | \quad \color{red}x_6 = 5\cdot x_2 + 3\cdot x_1\\ \mathbf{9} &\mathbf{=}& \mathbf{x_6} \\ \hline \end{array} \)

 

laugh

heureka  Oct 12, 2017
edited by heureka  Oct 12, 2017

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