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3.2 The sum of the first \( n \) terms of a sequence is: \[ S_{m}=3^{n-5}+2 \] Determine the \( 80^{\text {m }} \) term. Leave your answer in the form \( a . b^{p} \) where \( a, b \) and \( p \) are all integers

Ask by Weaver Black. in South Africa
Mar 16,2025

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The 80th term is \( 2 \times 3^{74} \).

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To find the \( n \)-th term \( T_n \) of the sequence, we use the relationship between the sum of the first \( n \) terms \( S_n \) and the \( n \)-th term \( T_n \): \[ T_n = S_n - S_{n-1} \] Given \( S_n = 3^{n-5} + 2 \), we first need to find \( S_{n-1} \): \[ S_{n-1} = 3^{(n-1)-5} + 2 = 3^{n-6} + 2 \] Now, let's calculate \( T_n \): \[ T_n = S_n - S_{n-1} = (3^{n-5} + 2) - (3^{n-6} + 2) \] Simplifying this, we have: \[ T_n = 3^{n-5} - 3^{n-6} \] Factoring out \( 3^{n-6} \): \[ T_n = 3^{n-6} (3 - 1) = 3^{n-6} \cdot 2 \] So, the \( n \)-th term can be expressed as: \[ T_n = 2 \cdot 3^{n-6} \] Now, we want to determine the \( 80^{\text{th}} \) term, which means setting \( n = 80 \): \[ T_{80} = 2 \cdot 3^{80-6} = 2 \cdot 3^{74} \] Thus the \( 80^{\text{th}} \) term is: \[ T_{80} = 2 \cdot 3^{74} \] In the required form \( a \cdot b^{p} \), we have: \[ a = 2, \quad b = 3, \quad p = 74 \] Therefore, the final answer is: \[ \boxed{2 \cdot 3^{74}} \]

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