MHB -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 sequence defined by the recursion a_(n+1) = a_n + n, with a_1 = -1, produces the first four terms as follows: a_1 = -1, a_2 = 0, a_3 = 1, and a_4 = 2. The calculations involve adding the current term to the index n, starting from the initial term. The process is iterative, with each subsequent term being derived from the previous one. The final result confirms that the first four terms of the sequence are -1, 0, 1, and 2.
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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

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

$\text{stuck!}$
 
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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
 
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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.

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
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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