U{n} = (-52n⁵ + 795n⁴ -4510n³ + 12045n² - 14518n + 6720)/120
Which gives U{1-5} = {4, 7, 13, 25, 49} and U6 = 42.
There are an infinite number of polynomials which will give the given sequence when the values 1-5 are input.
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What your teacher is most likely expecting is:
U{n} = 2.5 + 1.5 × (2ⁿ - 1)
or U{n} = (5 + 3 × (2ⁿ - 1)) / 2
Discovered via:
U2 = U1 + 3
U3 = U2 + 3 × 2
U4 = U3 + 3 × 2²
U{n} = U{n-1} + 3 × 2ⁿ⁻²
→ U1 = U0 + 3 × 2⁻¹
→ U0 = U1 - 3 × 2⁻¹
= 4 - 3/2
= 2.5
→ U1 = U0 + 3 × 2⁻¹ = 2.5 + 3 × 2⁻¹
U2 = U1 + 3 × 2⁰ = 2.5 + 3 × 2⁻¹ + 3 × 2⁰
U3 = U2 + 3 × 2¹ = 2.5 + 3 × 2⁻¹ + 3 × 2⁰ + 3 × 2¹
→ for n = 1, 2, 3, ...
U{n} = 2.5 + 3 × 2⁻¹ + 3 × 2⁰ + 3 × 2¹ + ... + 3 × 2ⁿ⁻²
= 2.5 + 3 × 2⁻¹ × (1 + 2¹ + 2² + ... + 2ⁿ⁻¹)
= 2.5 + 1.5 × (2ⁿ - 1)/(2 - 1)
= 2.5 + 1.5 × (2ⁿ - 1)
Assuming the sequence does not merely skip from 13 to 49, and instead carries on in the same pattern, the sequence proceeds thus:1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49.This is thirteen terms. The formula for finding these terms is 4x-3.
Ascending terms in the sequence are equal to 12, 32, 52, __, 92, and, for some reason, 102, not 112. Therefore, assuming the last term in the sequence to be 112 = 121, the fourth term in the sequence is 72 = 49.
81
3-5-7-9-11-13-15 answer to1-4-9-16-25-36 what are the next 2 numbers and fill space between these which is what we come up with 3-5-7-9-16-25-36-49-64
The difference between successive terms is 8 tn = 8n + 9 or 8(n+1)+1 where n = 1,2,3,...
Assuming the sequence does not merely skip from 13 to 49, and instead carries on in the same pattern, the sequence proceeds thus:1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49.This is thirteen terms. The formula for finding these terms is 4x-3.
Ascending terms in the sequence are equal to 12, 32, 52, __, 92, and, for some reason, 102, not 112. Therefore, assuming the last term in the sequence to be 112 = 121, the fourth term in the sequence is 72 = 49.
81
13/49 is in its lowest terms
121.
To find the nth term of a sequence, we first need to determine the pattern or rule that governs the sequence. In this case, the sequence appears to be increasing by adding consecutive odd numbers: 3, 6, 9, 12, and so on. Therefore, the nth term formula for this sequence is Tn = 3n^2 + n. So, the nth term for the sequence 4, 7, 13, 22, 34 is Tn = 3n^2 + n.
(Term)n = 59 - 2n
.... 49, 36, 25, 16, 9, 4 and 1
They are 25 and 49
To find the nth term of this sequence, we first need to identify the pattern. The differences between consecutive terms are 5, 9, 13, 17, and so on. These are increasing by 4 each time. This means that the nth term can be calculated using the formula n^2 + 4n + 1. So, the nth term for the sequence 5, 10, 19, 32, 49 is n^2 + 4n + 1.
49/25
3-5-7-9-11-13-15 answer to1-4-9-16-25-36 what are the next 2 numbers and fill space between these which is what we come up with 3-5-7-9-16-25-36-49-64