-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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The first four terms of the sequence defined by the recursion formula \( a_{n+1} = a_n + n \) with the initial term \( a_1 = -1 \) are calculated as follows: \( a_1 = -1 \), \( a_2 = 0 \), \( a_3 = 1 \), and \( a_4 = 2 \). The calculations proceed by substituting the previous term and the index into the recursive formula. Each term is derived sequentially, confirming the pattern established by the recursion.

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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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