Recent content by nightcrrawlerr

Dick, I do appreciate your response to my post. Your answers and explanations. You probably correct friend. Honestly I took this course twice already at first I drop this because I know I can't make it. For honest reason, I took this for the second time. Why? Because its part of my curriculum...

So dick pls. show how can i start and finish this up pls?

I'm just new about summation. They just give us this assignment. Kindly help me how can I get the correct answer on this. I only have 2 days to finish this up. So please...

What will the answer if i combine both equations? Is it -3 ^n^+^1

You mean my answer here is correct? Kindly help on this also, please. \sum^{\infty}_{k=0}\frac{k-1}{2^k} =\sum^n_{k=1}\frac{k}{2^k}-\sum^n_{k=0}\frac{1}{2^k} what should I do next?

Ok Dick and Tom Mattson, here i am again. kindly help me here out. Tom Mattson suggested this: \sum^n_{k=0}\frac{k-1}{2^k}=\sum^n_{k=1}\frac{k}{2^k}-\sum^n_{k=0}\frac{1}{2^k} How to solved the first and second equation? is it by geometric series still? thanks for helping me out.

Dick can i post here my answer for c.)? c.) \sum^n _{k=1} \frac{-1}{k(k + 1)} = \sum^n _{k=1} (\frac{-1}{k} + \frac {1}{k + 1}) = (\frac {-1}{1} + \frac {1}{2}) + (\frac {-1}{2} + \frac {1}{3}) + ... + (\frac{-1}{n} + \frac {1}{n + 1}) = {-1} + \frac{1}{n + 1}...

Guru, can i start posting my ans. on item c?

Ok sure you will. No problem for me. thanks

Based on the equation provided in question c. This is the correct equation. \sum^n _{k=1} {\frac {^-^1} _{k(k + 1)}} . Anyway I will verify this also... again thank you.

So you mean got the solution correct? how about the first term, how can i combine the answers in one equation? Kindly help me more, please. thank you.

for geometric series: = \sum^n _{k=0} a^k = a + a^1 + a^2 + ... + a^n for a "not equal" 1 = \sum^n _{k=0} 3^k = 3 + 3^1 + 3^2 + ... + 3^n multiply both sides with (1-a) (1 - a)\sum^n _{k=0} 3^k = \sum^n _{k=0} 3^k - \sum^n _{k=0} 3^k^+^1 = \sum^n _{k=0} 3^k - \sum...

for geometric series: = \sum^n _{k=0} a^k = a + a^1 + a^2 + ... + a^n for a "not equal" 1 = \sum^n _{k=0} 3^k = 3 + 3^1 + 3^2 + ... + 3^n multiply both sides with (1-a) (1 - a)\sum^n _{k=0} 3^k = \sum^n _{k=0} 3^k - \sum^n _{k=0} 3^k+^1 = \sum^n _{k=0} 3^k - \sum^n+^1...

Dick, thanks for you immediate answer. Ok i'll try first your comment on this and i will post it later on. Lets start first at letter a) \sum^n _{k=0} (3k - 3^k) this can be solved by telescoping series right? does this simplifying make sense sir? = 3 \sum^n _{k=0} k -...

Kindly comment with my answer. Kindly help me out of my answer please. If my answer is not correct kindly help me correct it out by teaching me, please. Thanks a lot for your great help guys. God bless and more power. for a. ) = \sum^n _{k=0} (3k - 3^k) = 3 \sum^n _{k=0} k -...