PMI and sequences

[Math3546]

Could someone help me?

 

[Deleted member 4993]

Could someone help me?

Please share your work/thoughts about this assignment.

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[Math3546]

Could someone help me?

I did most of these questions but not sure how to make it a constant. Do I just substitute with numbers? Also, I have some trouble with the last question iii) and iv ).

 

[Deleted member 4993]

I did most of these questions but not sure how to make it a constant. Do I just substitute with numbers? Also, I have some trouble with the last question iii) and iv ).

Please share your work/thoughts about this assignment.

 

[pka]

(N)ot sure how to make it a constant.

\(\displaystyle \large v_n=a+(n-1)d\) and \(\displaystyle \large u_n=\large2^{v_n}\)
\(\displaystyle \large{\dfrac{v_{n+1}}{v_n}=\dfrac{2^{a+nd}}{2^{a+(n-1)d}}=?}\)
Can you subtract exponents?

 

[Math3546]

\(\displaystyle \large v_n=a+(n-1)d\) and \(\displaystyle \large u_n=\large2^{v_n}\)
\(\displaystyle \large{\dfrac{v_{n+1}}{v_n}=\dfrac{2^{a+nd}}{2^{a+(n-1)d}}=?}\)
Can you subtract exponents?

Yeah, so it’s 2^-d and d doesn't equal to zero which means that 2^un is a constant

 

[Deleted member 4993]

Yeah, so it’s 2^-d and d doesn't equal to zero which means that 2^un is a constant

What is that "it"?

Can you write an equation showing - what is 2^-d ?

 

[pka]

Yeah, so it’s 2^-d and d doesn't equal to zero which means that 2^un is a constant

How did you get \(\displaystyle 2^{-d}~?\)
\(\displaystyle \Large{\dfrac{2^{a+nd}}{2^{a+(n-1)d}}=?}\)

 

[Math3546]

Ohh sorry, it’s actually 2^d. you do 2^a+dn /2^a+dn-d ,then we remove the parenthesis and we have 2^(a+dn) /2^(a+dn-d). We simplify the expansion and we get 2^d. Just wondering, do u have any idea how to do b) iii) and iv )?

 

[Math3546]

Ohh sorry, it’s actually 2^d. you do 2^a+dn /2^a+dn-d ,then we remove the parenthesis and we have 2^(a+dn) /2^(a+dn-d). We simplify the expansion and we get 2^d. Just wondering, do u have any idea how to do b) iii) and iv )?

I accidentally wrote everything in the power square but hopefully, you understand what I’ve meant

 

[pka]

I accidentally wrote everything in the power square but hopefully, you understand what I’ve meant

Because this has been kicking around, and I think it may be flawed, I will answer.
The sequence \(\displaystyle \Large{v_n=2^{a+(n-1)d}}\) that is part a) (iii).
For b) (i)\(\displaystyle \large{ S_m=\sum\limits_{n = 1}^m {{2^{a + (n - 1)d}}}=\dfrac{2^{a}(2^{md-1)})}{2^d-1}}\) SEE HERE

b) (iii) \(\displaystyle {S_\infty } = \sum\limits_{n = 1}^\infty {{2^{a + (n - 1)d}}} = - \frac{{{2^a}}}{{{2^d} - 1}}\text{ where }2^d<1\)

 

[Math3546]

Because this has been kicking around, and I think it may be flawed, I will answer.
The sequence \(\displaystyle \Large{v_n=2^{a+(n-1)d}}\) that is part a) (iii).
For b) (i)\(\displaystyle \large{ S_m=\sum\limits_{n = 1}^m {{2^{a + (n - 1)d}}}=\dfrac{2^{a}(2^{md-1)})}{2^d-1}}\) SEE HERE

b) (iii) \(\displaystyle {S_\infty } = \sum\limits_{n = 1}^\infty {{2^{a + (n - 1)d}}} = - \frac{{{2^a}}}{{{2^d} - 1}}\text{ where }2^d<1\)

This does actually make sense, thank you for clarifying.

 

[Math3546]

Could someone explain to me part a? I did something wrong, then I’ve corrected it, but still I want to make sure I’m right.