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The nth term of a sequence is defined as follows. Find the first four terms: (i) `T_(n)=3n+1` (ii) `T_(n)=n^(2)+5`

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To find the first four terms of the sequences defined by the nth terms \( T_n = 3n + 1 \) and \( T_n = n^2 + 5 \), we will substitute values of \( n \) from 1 to 4 into each formula. ### Part (i): \( T_n = 3n + 1 \) 1. **Calculate \( T_1 \)**: \[ T_1 = 3(1) + 1 = 3 + 1 = 4 \] 2. **Calculate \( T_2 \)**: \[ T_2 = 3(2) + 1 = 6 + 1 = 7 \] 3. **Calculate \( T_3 \)**: \[ T_3 = 3(3) + 1 = 9 + 1 = 10 \] 4. **Calculate \( T_4 \)**: \[ T_4 = 3(4) + 1 = 12 + 1 = 13 \] Thus, the first four terms of the sequence \( T_n = 3n + 1 \) are: \[ 4, 7, 10, 13 \] ### Part (ii): \( T_n = n^2 + 5 \) 1. **Calculate \( T_1 \)**: \[ T_1 = (1)^2 + 5 = 1 + 5 = 6 \] 2. **Calculate \( T_2 \)**: \[ T_2 = (2)^2 + 5 = 4 + 5 = 9 \] 3. **Calculate \( T_3 \)**: \[ T_3 = (3)^2 + 5 = 9 + 5 = 14 \] 4. **Calculate \( T_4 \)**: \[ T_4 = (4)^2 + 5 = 16 + 5 = 21 \] Thus, the first four terms of the sequence \( T_n = n^2 + 5 \) are: \[ 6, 9, 14, 21 \] ### Final Answers: - For \( T_n = 3n + 1 \): **4, 7, 10, 13** - For \( T_n = n^2 + 5 \): **6, 9, 14, 21**
To find the first four terms of the sequences defined by the nth terms \( T_n = 3n + 1 \) and \( T_n = n^2 + 5 \), we will substitute values of \( n \) from 1 to 4 into each formula. ### Part (i): \( T_n = 3n + 1 \) 1. **Calculate \( T_1 \)**: \[ T_1 = 3(1) + 1 = 3 + 1 = 4 \] ...
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