-z78 first four terms of the sequence of π‘Ž_(𝑛+1)=π‘Ž_𝑛+𝑛,π‘Ž_1=βˆ’1.

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Discussion Overview

The discussion revolves around determining the first four terms of a sequence defined by the recursive relation \( a_{n+1} = a_n + n \) with the initial condition \( a_1 = -1 \). Participants engage in calculating the terms step-by-step, exploring the implications of the recursion.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant attempts to compute the first term but expresses confusion about the recursion, stating \( a_{0+1} = -1 + 0 = -1 \).
  • Another participant correctly calculates the second term as \( a_2 = a_1 + 1 = -1 + 1 = 0 \) and invites others to continue the sequence.
  • Subsequent posts provide calculations for the third and fourth terms, with one participant stating the first four terms are \( -1, 0, 1, 2 \) based on their calculations.
  • Another participant revisits the calculations, suggesting that \( a_3 \) should be computed as \( a_2 + 2 = 0 + 2 = 2 \) and proposes that \( a_4 = a_3 + 3 = 2 + 3 = 5 \), indicating a potential discrepancy in the earlier terms presented.
  • There is a clarification about the recursive definition, emphasizing that \( a_1 \) is not a constant but is used in the calculations for subsequent terms.

Areas of Agreement / Disagreement

Participants generally agree on the recursive definition and the initial term, but there is disagreement regarding the correct values of the subsequent terms, leading to multiple competing views on the sequence's first four terms.

Contextual Notes

Some calculations appear to depend on the interpretation of the recursion, and there are unresolved discrepancies in the computed values for \( a_3 \) and \( a_4 \). The discussion reflects varying approaches to applying the recursive formula.

karush
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Write out the first four terms of the sequence defined by the recursion n_(n+1)=n_1+1,n_1=βˆ’1

$\text{Write out the first four terms of the sequence defined by the recursion}$
$$\displaystyle
a_{n+1}=a_1+1,a_1=βˆ’1$$.
$\text{so then}$
$$\displaystyle
a_{0+1}=-1+0=-1$$

$\text{stuck!}$
 
Last edited:
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Re: Write out the first four terms of the sequence defined by the recursion π‘Ž_(𝑛+1)=π‘Ž_𝑛+𝑛,π‘Ž_1=βˆ’1.

We are given the first term, so we need to manually compute the next 3 terms...here's the second one:

$$a_{1+1}=a_2=a_1+1=-1+1=0$$

Can you proceed?
 
Re: Write out the first four terms of the sequence defined by the recursion π‘Ž_(𝑛+1)=π‘Ž_𝑛+𝑛,π‘Ž_1=βˆ’1.

MarkFL said:
We are given the first term, so we need to manually compute the next 3 terms...here's the second one:

$$a_{1+1}=a_2=a_1+1=-1+1=0$$

Can you proceed?
$$a_{2+1}=a_3=a_1+1=-1+2=1$$
$$a_{3+1}=a_4=a_1+1=-1+3=2$$
$\textsf{so the first 4 terms are } $ $-1,0,1,2$
 
Re: Write out the first four terms of the sequence defined by the recursion π‘Ž_(𝑛+1)=π‘Ž_𝑛+𝑛,π‘Ž_1=βˆ’1.

karush said:
$$a_{2+1}=a_3=a_1+1=-1+2=1$$
$$a_{3+1}=a_4=a_1+1=-1+3=2$$
$\textsf{so the first 4 terms are } $ $-1,0,1,2$

We have:

$$a_1=-1$$

$$a_{1+1}=a_2=a_1+1=-1+1=0$$

$$a_{2+1}=a_3=a_2+2=0+2=2$$

$$a_{3+1}=a_4=a_3+3=2+3=5$$
 
Re: Write out the first four terms of the sequence defined by the recursion

so $a_1$ is not a constant but is replaced
 
Last edited:
Re: Write out the first four terms of the sequence defined by the recursion π‘Ž_(𝑛+1)=π‘Ž_𝑛+𝑛,π‘Ž_1=βˆ’1.

The recursive definition is:

$$a_{n+1}=a_{n}+n$$

So, when we compute $a_2$, we let $n=1$, and so only then will we have $a_1$ on the RHS of the definition. :D
 

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