Recent content by ChelseaL

So basically they are equal?

Use the fact that \frac{1}{k} - \frac{1}{k+1} = \frac{1}{k(k+1)} to show that Deduce that \infty sigma (\frac{1}{k(k+1)}) = 1 r=1 How do I solve this?

Thank you!

This is my first time hearing this term. I don't think my lecturer expects us to use that method since we never covered it in class.

Sn = 1 + (1/n+1)?

How do I work it out from there?

Use the fact that \frac{1}{k} - \frac{1}{k+1} = \frac{1}{k(k+1)} to show that n sigma (\frac{1}{k(k+1)}) = 1- \frac{1}{n+1} r=1 What do I need to do to solve it?

I'm not exactly sure what the topic is, but this is the part before it. https://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/geometric-series-24025.html

Given that the sum of the first n terms of series, s, is 9-32-n Find the sum of infinity of s. Do I use the formula S\infty = \frac{a}{1-r}?

ohhhh. What do I need to do from here?

\frac{1}{n+1}?

18/3n + 1 / (18/3n)? Sorry for not really understanding this topic. My lecturer regularly misses class and refuses to help us when we ask questions.

What do you mean?

So far I have this 9-32-n 3 = 9 = 9 -(n-2) = 9 -(n-1-1) = 9 -(n-1) + 1 = 9 \cdot 3 -(n-1) = 9 (\frac{1}{3}) n-1 where a=9 and r=\frac{1}{3}

Given that the sum of the first n terms of series, s, is 9-32-n show that the s is a geometric progression. Do I use the formula an = ar n-1? And if so, how do I apply it?