A sequence is (recursively) defined as $a_{1} = 0$ $a_{2} = 1$ and for $n > 2$ $a_{n} = a_{n - 1} - 3 a_{n - 2}$ What is $a_{6}$ ?

- 8

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\frac{19}{6}

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18

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72

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109

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Solution

The correct option is B $\frac{19}{6}$
Given that $a_{1} = 0 a_{2} = 1$
$2 a_{n} = a_{n - 1} - 3 a_{n - 2} 2 a_{3} = a_{2} - 3 a_{1} \Rightarrow a_{3} = \frac{1 - 0}{2} = \frac{1}{2} 2 a_{4} = a_{3} - 3 a_{2} \Rightarrow a_{4} = \frac{\frac{1}{2} - 3}{2} = \frac{- 5}{4} 2 a_{5} = a_{4} - 3 a_{3} \Rightarrow a_{5} = \frac{1}{2} [\frac{- 5}{4} - \frac{3}{2}] = \frac{- 11}{8} 2 a_{6} = a_{5} - 3 a_{4} \Rightarrow a_{6} = \frac{1}{2} [\frac{- 11}{8} - 3 (\frac{- 5}{4})] = \frac{19}{16}$

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A sequence is (recursively) defined as a1=0 a2=1 and for n>2 an=an−1−3an−2 What is a6?

The correct option is B 196Given that a1=0a2=12an=an−1−3an−22a3=a2−3a1⇒a3=1−02=122a4=a3−3a2⇒a4=12−32=−542a5=a4−3a3⇒a5=12[−54−32]=−1182a6=a5−3a4⇒a6=12[−118−3(−54)]=1916

A sequence is (recursively) defined as $a_{1} = 0$ $a_{2} = 1$ and for $n > 2$ $a_{n} = a_{n - 1} - 3 a_{n - 2}$ What is $a_{6}$ ?