What is the Formula for the Sequence 242.9.1.24-26?

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$\text{write the formula for the sequence}$
$\text{.24 }$ $$a_n=\frac{n^{n+1}}{5^{n+1}}$$
$\text{.25 }$ $$a_n=\frac{(-1)^{n}+1}{2}$$
$\text{26.}?$
 
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karush said:
$\text{write the formula for the sequence}$
$\text{.24 }$ $$a_n=\frac{n^{n+1}}{5^{n+1}}$$
$\text{.25 }$ $$a_n=\frac{(-1)^{n}+1}{2}$$
$\text{26.}?$

Is 4^5 really 64? Is 5^6 really 125?
 
Prove It said:
Is 4^5 really 64? Is 5^6 really 125?
$\text{.24 }$ $$a_n=\frac{n^3}{5^{n+1}}$$
 
For problem 25, I would consider using the magnitude or the square of a trigonometric function. :D
 
MarkFL said:
For problem 25, I would consider using the magnitude or the square of a trigonometric function. :D

While that would work, I think it would complicate things :P why note just a simple hybrid function?

$\displaystyle \begin{align*} a_n = \begin{cases} 1 \textrm{ if } n \textrm{ is odd} \\ 0 \textrm{ if } n \textrm{ is even} \end{cases} , \, n \in \mathbf{N} \end{align*}$
 
For Q3 I would do $\displaystyle \begin{align*} a_n = \left \lceil {\frac{n}{2}} \right \rceil , \, n \in \left\{ 0, 1, 2, 3, \dots \right\} \end{align*}$
 
$\text{26 the sequence 0, 1, 1, 2, 2, 3, 3, 4····}$
$$\displaystyle
a_n=a_n +\frac{(-1)^n +1}{2} \\
a_1=0\\
a_2=0+\frac{(-1)^2 +1}{2}=1 \\
a_3=1+\frac{(-1)^3 +1}{2}=1 \\
a_4=1+\frac{(-1)^4 +1}{2}=2 \\
a_5=2+\frac{(-1)^5 +1}{2}=2 \\
a_5=2+\frac{(-1)^5 +1}{2}=3 \\
a_5=3+\frac{(-1)^5 +1}{2}=3 \\
a_5=3+\frac{(-1)^5 +1}{2}=4 \\
$$
$\text{no given answer on this so hopefully..}$
☕
 
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Prove It said:
For Q3 I would do $\displaystyle \begin{align*} a_n = \left \lceil {\frac{n}{2}} \right \rceil , \, n \in \left\{ 0, 1, 2, 3, \dots \right\} \end{align*}$

how?
 
karush said:
how?
Do you know what the notation means? The "ceiling function", [tex]\lceil {x}\rceil[/tex] (this is not "[x]"- there are no lower serifs), is equal to the smallest integer less than or equal to x. So [tex]\left\lceil 0\right\rceil= 0[/tex], [tex]\left\lceil \frac{1}{2}\right\rceil= 0[/tex], [tex]\left\lceil 1\right\rceil= 1[/tex], [tex]\left\lceil \frac{3}{2}\right\rceil= 1[/tex], etc.
 
HallsofIvy said:
Do you know what the notation means? The "ceiling function", [tex]\lceil {x}\rceil[/tex] (this is not "[x]"- there are no lower serifs), is equal to the smallest integer less than or equal to x. So [tex]\left\lceil 0\right\rceil= 0[/tex], [tex]\left\lceil \frac{1}{2}\right\rceil= 0[/tex], [tex]\left\lceil 1\right\rceil= 1[/tex], [tex]\left\lceil \frac{3}{2}\right\rceil= 1[/tex], etc.

It's actually the smallest integer GREATER THAN or equal to x...