Computing terms yields 1, 2, 4, 8, 16, 32, 64, 128. Since Pn – Pn-1 = Pn-1, the simpler recursive rule is Pn = 2Pn-1 for n >= 3. The explicit formula is Pn = 2ⁿ⁻¹ for n >= 2 ...
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With a = 2 and r = 4, explicit formula is tn = 2 x 4ⁿ⁻¹ and recursive formula is t1 = 2, tn = 4tn-1 for n >= 2. Solving 2 x 4ⁿ⁻¹ = 131072 gives 4ⁿ⁻¹ = 65536 ...
The sum of k consecutive natural numbers is k(2a + k – 1) / 2 = 100, meaning k(2a + k – 1) = 200. Testing odd factors gives only k = 5, yielding 18 + 19 + 20 + ...
Given a + 10d = 38 and a + 15d = 73, subtracting gives 5d = 35, so d = 7 and a = -32. Using tn = a + (n – 1)d, the 31st term is -32 + 30 ...
Starting with t1 equal to 2, the recursive rule gives the sequence 2, 4, 10, 28, 82, 244, 730. Therefore, 730 occurs at the seventh position. Hence, 730 is the seventh term of the sequence.