Geometric Series in Sigma Notation with natural log

Question

So I am trying to solve for this problem below, I attach a screenshot of it, but I just can't seem to find the answer. Is anyone able to solve this? (Extra note: U_n = U_r)

So for (a) I got 2(1-2^-10), which i got from deriving the geometric series equation $$S_n = \frac{U_1(1-r^n)}{r-1}$$ Which I believe should be correct, but for question (b) I tried to change the sigma notation series $$\sum_{n=0}^{10} ln(0.5^n)$$ into the normal geometric series equation $$S_{10} = \frac{1(1-ln(0.5^{10}))}{1-ln(0.5^{10})}$$ But then I got different values for it. So, is there another method to get the exact value in the form of c ln d, where c and d is an integer? And is my answer for question (a) correct?

See the problem here

Answer 1

(a) Given sum is nothing else than:

$\sum_{r=0}^{10}u_r=u_0+u_1+u_2+u_3+u_4+u_5+u_6+u_7++u_8+u_9+u_{10}$.

Next, we have $u_n=0.5^n$, so we have to find:

$0.5^0 + 0.5^1 + 0.5^2 + 0.5^3 + 0.5^4 + 0.5^5 + 0.5^6 + 0.5^7 + 0.5^8 + 0.5^9 + 0.5^{10}$.

It is easy to recognize that the given sequence $u_n$ is geometric sequence, so basically we have to find the sum of first $11$ members of it. To find such sum we can use well known formula, i.e.

$S_{11} = \frac{1 - (\frac{1}{2})^{11}}{1-\frac{1}{2}} = ... = \frac{2047}{1024} = \frac{2048-1}{2^{10}} = \frac{2^{11} - 2^0}{2^{10}}$.

Answer 2

Hint: The sum in part (b) is not a geometric series. But we do know that $\ \ln\left(a^n\right)=n\ln(a)\ $ for any positive $\ a\ $, so $\ \sum_\limits{n=0}^{10}\ln\left(0.5^n\right)= \sum_\limits{n=0}^{10}n\ln(0.5)\ $. Can you now work out the sum of this series?