# Thread:

1. ## Sequences and Series + Natural Log

I've spent so long trying to figure this out and I just CAN'T. I can see the relationship but I don't know how to actually calculate this through

Sn = u1(r^n-1) / 4-1
or Sn = n/2(2u1 + (n-1)d)
I tried calculating them both as geometric and as arithmetic but I can't figure out a consistent d or r?

Thank you.

2. Originally Posted by positiveion

I've spent so long trying to figure this out and I just CAN'T. I can see the relationship but I don't know how to actually calculate this through

Sn = u1(r^n-1) / 4-1
or Sn = n/2(2u1 + (n-1)d)
I tried calculating them both as geometric and as arithmetic but I can't figure out a consistent d or r?

Thank you.
For this sequence to be arithmetic each term must be separated by a common difference. In other words $U_3 - U_2 = U_2 - U_1$

Since $\left(\frac{[\ln(2)]^3}{3!} - \frac{[\ln(2)]^2}{2!}\right) \neq \left(\frac{[\ln(2)]^2}{2!} - \frac{[\ln(2)]}{1!}\right)$ then the series is not arithmetic

More it to be geometric then the ratio of each successive term must be equal:

$\frac{U_3}{U_2} = \frac{U_2}{U_1}$

$\left(\frac{(\ln2) ^3}{3!} \times \frac{2!}{(\ln2) ^2}\right) \neq \left(\frac{(\ln2) ^2}{2!} \times \frac{1!}{\ln2}\right)$ so it is not geometric either.

From what I can gather this is of the form $\frac{(\ln2)^n}{n!}$ but I do not know where to go from there but hopefully someone else may be able to take it further

edit: are you told how x and a relate to the sequence?

3. I recognize this as the Maclaurin series for $f(x) = a^x$ with $a = 2$ and $x =1$, but I'm not sure how this might help you to solve.

4. ahgh i feel so confused

i print screened more of the worksheet

does this additional info make any difference?
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