Can anyone tell me the nth term of the sequence 2,3,5,8,12,17?

i know that the number added keeps increasing by one.

but what is the nth term of the sequence???

15 Answers

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  • 1 decade ago
    Best Answer

    The answer you want is:

    a(n) = (n² - n + 4) / 2

    I will do more than tell you, or even explain what the sequence is, I will tell you how to do the question. Nobody seems too intent on explaining how you would arrive at the answer (and more than half the answers don't even have the right sequence, and almost nobody's giving you the equation for it, just the next few numbers).

    Whenever you're looking at a sequence, one of the first things to look at is the differences in the sequence. They form their own sequence:

    3 - 2 = 1

    5 - 3 = 2

    8 - 5 = 3

    12 - 8 = 4

    17 - 12 = 5

    You can even do it again, with this "diff-sequence"

    2 - 1 = 1

    3 - 2 = 1

    4 - 3 = 1

    5 - 4 = 1

    The goal is to repeat this process until you get down to the same number. If this is possible (as it has happened here, they are all 1s), then you're in luck. The differences all skip by 1s, so the first difference is 1, the second is 2, etc. So you can say:

    d(n) = n

    The sequence a(n) is defined by those differences as:

    a(n+1) = a(n) + d(n) = a(n) + n

    This is called a recursive formula, because to get the (n+1)th term you need the nth term. The next few terms are easy to calculate but..

    ...how can you find the formula without this? If I asked for the 1000th term, you'd have to calculate all 1000 of them!! That's too much work.

    To figure this out, look at the process of "diff-sequences" above. Because it took two steps to get to a repeating pattern (1,1,1,1...), this is a polynomial of degree 2.

    So let's say:

    a(n) = A n² + B n + C

    How do we find A, B, C? Well, there are three ways I can think of. One way is to use calculus. If you do not know calculus, skip to the second method below.

    ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- -----

    METHOD 1

    For this to work, you have to consider a(0)=2, a(1)=3, etc. Then clearly C=2, so that's taken care of. The variable that we differentiate here is n, not x.

    Start with:

    a(n) = An² + Bn + C

    a'(n) = 2 A n² + B

    a''(n) = 2A

    Now, a''(n) is the second diff-sequence, meaning:

    a''(n) = 1

    a''(n) = 2A = 1

    So A = 1/2.

    Finally, just use a(1) to find B:

    a(1) = (1/2)(1)² + B(1) + 2 = 3

    1/2 + B + 2 = 3

    B = 3 - 5/2 = 1/2

    We cannot use a'(n) to find B because a'(n) is not the same as the second diff-sequence (1,2,3,4,5). The second diff-sequence is offset due to the fact that this is a sequence, not a function. Only the constant diff-squence (1,1,1,1) is of use.

    So:

    a(n) = (1/2)n² + (1/2)n + 2

    Remember - this is different then method 2. It will give you the sequence 2,3,5,8 for n=0,1,2,3. To correct this, simply change all the ns to (n-1)s:

    a(n) = (1/2)(n-1)² + (1/2)(n-1) + 2

    this simplifies to the answer you will get from method 2:

    a(n) = (1/2)n² - (1/2)n + 2 = (n² - n + 4)/2

    ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- -----

    METHOD 2

    The algebraic method is pretty standard. You have at least three terms of your sequence so you can plug in three values of n and figure out the three coefficients:

    a(1) = A + B + C = 2 [EQ1]

    a(2) = 4A + 2B + C = 3 [EQ2]

    a(3) = 9A + 3B + C = 5 [EQ3]

    So solving these equations is our goal. So take EQ2 - EQ1 and EQ3 - EQ1:

    3A + B = 1 [EQ4]

    8A + 2B = 3 [EQ5]

    Then subtract: EQ5 - 2*EQ4

    2A = 1

    A = 1/2

    Substitute that into [EQ4]:

    3(1/2) + B = 1

    B = -1/2

    Plug A and B back into EQ1

    (1/2)+(-1/2) + C = 2

    C = 2

    So a(n) = (1/2)n² + (-1/2)n + 2

    which can be better written:

    a(n) = (n² - n + 4) / 2

    ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- -----

    METHOD 3

    The last method is a method called "generating functions." This is a more advanced method of solving recursive equations that's probably not worth writing out here. I'll refer everyone to:

    http://en.wikipedia.org/wiki/Generating_function

    Edit: On second thought, since I have the time, I will do it. If you do not follow this method, then just ignore it. Method 1 or 2 above is fine.

    Now, like method 1, we will have to start with a(0) instead of a(1). To make it work out like we want though, we can just do a(0)=2, a(1) = a(0)+0 = 2+0 = 2.

    We have:

    a(n+1) = a(n) + n

    Σ[n=0 to ∞] a(n+1) x^(n+1) = x Σ[n=0 to ∞] a(n) x^n + x Σ[n=0 to ∞]n x^n

    Let F(x) = Σ[n=0 to ∞] a(n)x^n

    Thus:

    Σ[n=1 to ∞] a(n) x^n = x F(x) + x Σ[n=0 to ∞]n x^n

    F(x) - a(0) = x F(x) + x² Σ[n=1 to ∞]n x^(n-1)

    Now we turn our attention to:

    Σ[n=0 to ∞]n x^(n-1) = (d/dx) Σ[n=0 to ∞] x^n

    But Σ[n=0 to ∞] x^n = 1/(1-x)

    (d/dx) [ 1/(1-x) ] = 1/(1-x)²

    Thus:

    Σ[n=0 to ∞]n x^(n-1) = 1/(1-x)²

    Finally, put that back into the equation with F(x):

    F(x) - a(0) = x F(x) + x² / (1-x)²

    Substitute a(0)=2

    F(x) - 2 = x F(x) + x²/(1-x)²

    (1-x) F(x) = x²/(1-x)² + 2

    F(x) = x² / (1-x)³ + 2/(1-x)

    Lastly, we need the Taylor series:

    1/(1-x)³ = Σ[n=0 to ∞] [n(n+1)/2] x^(n-1)

    1/(1-x) = Σ[n=0 to ∞] x^n

    F(x) = x²Σ[n=0 to ∞] [ n(n-1)/2] x^(n-1) +2Σ[n=0 to ∞] x^n

    F(x) = Σ[n=0 to ∞] [n(n-1)/2] x^(n+1) + 2Σ[n=0 to ∞] x^n

    F(x) = Σ[n=0 to ∞] [n(n-1)/2] x^(n) + 2Σ[n=0 to ∞] x^n

    F(x) = Σ[n=0 to ∞] ( n(n-1)/2 + 2 ) x^(n)

    F(x) = Σ[n=0 to ∞] (n²-n+4)/2 x^(n)

    Σ[n=0 to ∞] a(n)x^n = Σ[n=0 to ∞] (n²-n+4)/2 x^(n)

    And finally drop the series to get:

    a(n) = (n²-n+4)/2

    Exactly what the other methods determined. This is a more complicated (and more advanced) method than you need for this problem, but methods 1 and 2 do not always work, which means for some other problems, this method is the only method.

    Source(s): I am a professional mathematician who studies this sort of thing. Trust me, this is how you do it. If none of this makes sense but method 2, use method 2. Whoever gave this a thumbs down is a really misguided person and a poor mathematcian.
  • Erika
    Lv 4
    3 years ago

    Nth Term Of A Sequence

  • 1 decade ago

    Differences between the terms: 1,2,3,4,5,.......

    M(n) = M(n-1) + (n-1)

    Look at some even terms:

    M(4) = 2 + (n-1) + (n-2) + (n-3) = 2 + 3(n - 2) and 2 = n/2 / 3 = n-1

    M(6) = 2 + 5n - 15 = 2 + 5(n - 3) and again 3 = n/2 / 5 = n-1

    So in general:

    2 + (n - 1)(n - n/2) = 2 + n(n-1)(1/2)

    Now some odd terms:

    M(3) = 2 + (n - 1) + (n - 2) = 2 + 2n - 3

    M(5) = 2 + (n - 1) + (n - 2) + (n - 3) + (n - 4) = 2 + 4n - 10

    2 + 2n - 3 = 2 + (n-1)n - 3 and 3 = n(n-1)/2 / 2 = n - 1

    2 + (n - 1)n - 10 and 10 = n(n-1)/2 / 4 = n - 1

    So these also give:

    2 + n(n-1)(1/2) as the general expression

    Using this expression we get:

    1 2

    2 3

    3 5

    4 8

    5 12

    6 17

    Which matches the original sequence.

  • Dale
    Lv 4
    1 decade ago

    Consider successive differences:

    3-2=1

    5-3=2

    8-5=3

    12-8=4

    17-12=5

    Hence the next difference is 6 and so

    17+6=23 is the next term.

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  • 1 decade ago

    2

    2 + 1=3

    3 + 2=5

    5 + 3=8

    8 + 4=12

    12 + 5 = 17

    And....

    17 + 6 = 23!

  • 1 decade ago

    23

  • 1 decade ago

    Use the formula

    a + (n-1)d + 0.5(n-1)(n-2)C

    Where -

    d = the first difference (1 in the sequence).

    C = The difference increase. (In this case, again, 1)

    a = the first term.

  • 1 decade ago

    Its a Fibonacci sequence. Each term except first two, is a sum of the previous two terms. Now try it!

    But do not look for a formula for the nth term. You may spend days and days and yet may not get it.

  • Anonymous
    1 decade ago

    is nth like the number of terms that it is into the sequence? so like 2 is the first number and 3 is the second number (well, that bits kind of obvious)

    so you just have to relate it to n (which i cant do)

    look at how they are all similar to the number of sequence they are and try to find how they are common

  • 1 decade ago

    /

    nth term=n^2/2-n/2+2

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