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1 2 5 14 41

Since every term is expressed with respect to the immediately preceding term, this is a recursive sequence, where t1 = 1 and tn+1 = 3tn - 1, n = 1, 2, 3, ...

t1 = 1

t2 = 3(1) - 1 = 2

t3 = 3(2) - 1 = 5

t4 = 3(5) - 1 = 14

t5 = 3(15) - 1 = 41

next

t6 = 3(41) - 1 = 122

Q: What is the sequence 1 2 5 14 41?

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41 should be in between 14 and 122

The sequence is xn = xn-1 + 2

The next number in the sequence is 15 - the sequence goes x2, +2, -1

The simplest sequence appears to be 15 1*2=2 2+2=4 4-1=3 3*2=6 6+2=8 8-1=7 7*2=14 14+2=16 16-1=15 Multiply by 2 Add 2 Subtract 1 Repeat

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The sequence has the formula is 3x - 1 2 → 6 - 1 = 5 5 →15 - 1 = 14 14 →42 -1 = 41 41 → 123 - 1 = 122 The missing number is therefore 41.

Type your answer here... The sequence has the formula is 3x - 1 2 → 6 - 1 = 5 5 →15 - 1 = 14 14 →42 -1 = 41 41 → 123 - 1 = 122 The missing number is therefore 41.

41 122 365

41 should be in between 14 and 122

2 (+3) 5 (+9) 14 (+27) 41 (+81) 122 in this series... the difference is multiplied by 3. like 3,9,27,81,243...

The sequence is xn = xn-1 + 2

The next number in the sequence is 15 - the sequence goes x2, +2, -1

15 The sequence seems to be x2, +2, -1.

Add 3n-1. For example 1 + 30 = 2 2 + 31 = 5 5 + 32 = 14 14 + 33 = 41

The simplest sequence appears to be 15 1*2=2 2+2=4 4-1=3 3*2=6 6+2=8 8-1=7 7*2=14 14+2=16 16-1=15 Multiply by 2 Add 2 Subtract 1 Repeat

x 2 + 2 - 1 is the repeating sequence 1 x 2 = 2 2 + 2 = 4 4 - 1 = 3 3 x 2 = 6 6 + 2 = 8 8 - 1 = 7 7 x 2 = 14 14 + 2 = 16 16 - 1 = 15 the next number is 15

The simplest sequence appears to be 15 1*2=2 2+2=4 4-1=3 3*2=6 6+2=8 8-1=7 7*2=14 14+2=16 16-1=15 Multiply by 2 Add 2 Subtract 1 Repeat