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
49/25
Ah, what a delightful sequence you have there, friend! That sequence is called the "square numbers sequence." Each number is a perfect square - the result of multiplying a number by itself. Keep exploring the beauty of numbers and patterns, and let your creativity flow like a happy little stream.
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