I view the sequence has actually a team of 3 terms that repeats: twin first term come get 2nd term, add three come get third term, repeat.

What about: $2, 3, 6, 7, 14, 15, 30,...$?

Again the sequence has a team of three terms the repeats: add one to very first term come get 2nd term, then dual second term to get third term.

How do you compute the $n$th term of together sequences directly, without iterating through all preceding terms?

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edited Jul 21 "14 in ~ 20:30 hardhunterriverpei.com
inquiry Jul 20 "14 at 11:13 GuestGuest
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For the first sequence:

$$S_n = 2^(2+\lfloor\fracn2\rfloor)+6(\fracn2-\lfloor\fracn2\rfloor-1)$$

For example:

$S_1 = 2^2-3 = 1$$S_2 = 2^3-6 = 2$$S_3 = 2^3-3 = 5$$S_4 = 2^4-6 = 10$$S_5 = 2^4-3 = 13$$S_6 = 2^5-6 = 26$$S_7 = 2^5-3 = 29$

For the 2nd sequence:

$$S_n = 2^\lceil\fracn2\rceil+2(\fracn2-\lfloor\fracn2\rfloor-1)$$

For example:

$S_1 = 2^2-2 = 2$$S_2 = 2^2-1 = 3$$S_3 = 2^3-2 = 6$$S_4 = 2^3-1 = 7$$S_5 = 2^4-2 = 14$$S_6 = 2^4-1 = 15$$S_7 = 2^5-2 = 30$
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edited Jul 20 "14 at 12:23
answered Jul 20 "14 at 11:43 barak manosbarak manos
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For example, think about the first sequence. One have the right to write 2 recurrence relations,$$F_2k=2F_2k-1,\qquad F_2k+1=F_2k+3,$$and usage them come deduce a relation entailing only strange terms:$$F_2k+1=2F_2k-1+3.$$The basic solution of this is$$F_2k+1=\alpha\cdot 2^k-3,$$and the value of the constant $\alpha=2$ is solved by $F_1=1$. Hence$$F_2k+1=2^k+2-3,\qquad F_2k=2^k+2-6.$$

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edited Jul 20 "14 at 18:09
reply Jul 20 "14 in ~ 11:45 start wearing purpleStart wearing purple
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Denote the $n$-th ax by $x_n$ and have a look in ~ the state withodd table of contents $n$. Based upon what friend remarked yourself:

$x_1=1$

$x_3=2x_1+3=2+3$

$x_5=2x_3+3=2\left(2+3\right)+3=2^2+2.3+3$

$x_7=2x_5+3=2\left(2^2+2.3+3\right)+3=2^3+2^2.3+2.3+3$

et cetera.

You are watching: 1 - 2 - 5 - 10 - 13 - 26 - 29 - 48

This starts to look like

$x_2n+1=2^n+\left(2^n-1+2^n-2+\cdots+1\right)3=2^n+\left(2^n-1\right)3=2^n+2-3$

This have the right to be confirmed by induction and also its easy now to uncover that $x_2n=x_2n+1-3=2^n+2-6$.

On a sortlike means you concerned $y_2n+1=2^n+2-2$ and $y_2n=2^n+1-1$ whereby $y_n$denotes the $n$-th ax of the second sequence.

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edited Jul 20 "14 at 12:02
reply Jul 20 "14 in ~ 11:49 drhabdrhab
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A much more general method to addressing for the $n$th term of such sequences supplies matrix multiplication. Intend the even terms are nonzero consistent $r\neq 1$ time the coming before odd terms, when the odd state are constant $d$ to add the preceding even terms. Us have:

$$\beginpmatrix 1 & 0 & d \\ 1 & 0 & 0 \\ 0 & 0 & 1 \endpmatrix \beginpmatrix a_2n \\ a_2n-1 \\ 1 \endpmatrix = \beginpmatrix a_2n+1 \\ a_2n \\ 1 \endpmatrix$$

$$\beginpmatrix r & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \endpmatrix \beginpmatrix a_2n+1 \\ a_2n \\ 1 \endpmatrix = \beginpmatrix a_2n+2 \\ a_2n+1 \\ 1 \endpmatrix$$

Combining these by matrix multiplication offers the twin step:

$$\beginpmatrix r & 0 & rd \\ 1 & 0 & d \\ 0 & 0 & 1 \endpmatrix \beginpmatrix a_2n \\ a_2n-1 \\ 1 \endpmatrix = \beginpmatrix a_2n+2 \\ a_2n+1 \\ 1 \endpmatrix$$

The trouble is then decreased to recognize a closed form for natural powers of the matrix:

$$A = \beginpmatrix r & 0 & rd \\ 1 & 0 & d \\ 0 & 0 & 1 \endpmatrix$$

which have the right to be excellent by diagonalization, due to the fact that $A$ has three distinctive eigenvalues $0,1,r$.

Represent $A$ v respect come the equivalent basis of eigenvectors:

$$\left\ \beginpmatrix 0 \\ 1 \\ 0 \endpmatrix, \beginpmatrix -rd \\ - d \\ r-1 \endpmatrix, \beginpmatrix r \\ 1 \\ 0 \endpmatrix \right\$$

and the resulting similarity transformation diagonalizes $A$, say $A = S D S^-1$ whereby $D= \operatornamediag(0,1,r)$. Thus, assuming an initial worth $a_1$ and also $a_2 = ra_1$:

$$A^n \beginpmatrix a_2 \\ a_1 \\ 1 \endpmatrix = S D^n S^-1 \beginpmatrix ra_1 \\ a_1 \\ 1 \endpmatrix =\beginpmatrix a_2n+2 \\ a_2n+1 \\ 1 \endpmatrix$$

The strength of $D$ are clearly $D^n = \operatornamediag(0,1,r^n)$, for this reason this provides a direct expression for any terms in the sequence beginning from $a_1$:

$$a_2n+1 = r^n a_1 + \fracr^n -1r-1 d$$

$$a_2n+2 = r a_2n+1 = r^n+1 a_1 + \fracr^n -1r-1 rd$$

taking benefit of the calculate DanielV lugged out in the Comment below.

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This procession multiplication an approach can it is in modified to handle much more general mixtures of arithmetic and also geometric rules.