Standard +0.3 This is a straightforward proof by induction of a recurrence relation formula with a clear structure. The base case is trivial (u₁=1), and the inductive step requires substituting the recurrence relation into the assumed formula and simplifying using basic algebra and index laws. This is a standard textbook exercise requiring routine application of the induction framework with no novel insight or tricky algebraic manipulation.
6.
A sequence of positive integers \(u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots\) is defined by \(\left\{ \begin{array} { c } u _ { 1 } = 1 \\ u _ { n + 1 } = 3 u _ { n } + 2 \end{array} ( n \geq 1 ) \right.\)
Prove by induction that \(u _ { n } = 2 \left( 3 ^ { n - 1 } \right) - 1\).
(4) [0pt]
6.
A sequence of positive integers $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$ is defined by $\left\{ \begin{array} { c } u _ { 1 } = 1 \\ u _ { n + 1 } = 3 u _ { n } + 2 \end{array} ( n \geq 1 ) \right.$\\
Prove by induction that $u _ { n } = 2 \left( 3 ^ { n - 1 } \right) - 1$.\\
(4)\\[0pt]
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\hfill \mbox{\textit{SPS SPS FM 2022 Q6 [4]}}