gaokao 2018 Q17

gaokao · China · national-I-arts 12 marks Geometric Sequences and Series Geometric Sequence from Recurrence Identification

Given a sequence $\{ a _ { n } \}$ satisfying $a _ { 1 } = 1$ and $n a _ { n - 1 } = 2 ( n + 1 ) a _ { n }$. Let $b _ { n } = \frac { a _ { n } } { n }$.
(1) Find $b _ { 1 } , b _ { 2 } , b _ { 3 }$;
(2) Determine whether the sequence $\{ b _ { n } \}$ is a geometric sequence and explain the reasoning;
(3) Find the general term formula for $\{ a _ { n } \}$.

(1) From the given condition, $a _ { n + 1 } = \frac { 2 ( n + 1 ) } { n } a _ { n }$. Substituting $n = 1$, we get $a _ { 2 } = 4 a _ { 1 }$. Since $a _ { 1 } = 1$, we have $a _ { 2 } = 4$. Substituting $n = 2$, we get $a _ { 3 } = 3 a _ { 2 }$, so $a _ { 3 } = 12$. Thus $b _ { 1 } = 1 , b _ { 2 } = 2 , b _ { 3 } = 4$.
(2) $\{ b _ { n } \}$ is a geometric sequence with first term 1 and common ratio 2. From the condition, $\frac { a _ { n + 1 } } { n + 1 } = \frac { 2 a _ { n } } { n }$, so $b _ { n + 1 } = 2 b _ { n }$. Since $b _ { 1 } = 1$, $\{ b _ { n } \}$ is a geometric sequence with first term 1 and common ratio 2.
(3) From (2), $\frac { a _ { n } } { n } = 2 ^ { n - 1 }$, so $a _ { n } = n \cdot 2 ^ { n - 1 }$.

Given a sequence $\{ a _ { n } \}$ satisfying $a _ { 1 } = 1$ and $n a _ { n - 1 } = 2 ( n + 1 ) a _ { n }$. Let $b _ { n } = \frac { a _ { n } } { n }$.\\
(1) Find $b _ { 1 } , b _ { 2 } , b _ { 3 }$;\\
(2) Determine whether the sequence $\{ b _ { n } \}$ is a geometric sequence and explain the reasoning;\\
(3) Find the general term formula for $\{ a _ { n } \}$.

Paper Questions

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