# Simplifying summations

by nightcrrawlerr
Tags: simplifying, summations
Emeritus
PF Gold
P: 5,534
 Quote by nightcrrawlerr Still I couldn't figure it out. By the way what's RHS?
Hehe, sorry.

RHS=Right Hand Side
LHS=Left Hand Side

The only part of this that wasn't answered as of my last post, has now been answered by Dick. Can you be specific about what is still giving you trouble?

 Could you recomend some reading, books, website links, and etc. to further my study of this, please.
I am drawing only on my knowledge of what we here in the States call "Calculus II".
Emeritus
PF Gold
P: 5,534
 Quote by Dick $$\sum^n_{k=0}(\frac{x}{2})^k= \frac{1-(\frac{x}{2})^{n+1}}{1-(\frac{x}{2})}$$ Right? Assuming I did the tex ok. Differentiate, put x=1. The LHS is the tough part of a). The RHS is the answer.
OK, I understand you now. The part that confused me was when you said to use a "power series". What you have posted above is not a power series, because it doesn't have infinitely many terms.

Though I suppose you could turn it into a power series by letting the index run to infinity and multiplying the summand by a coefficient $a_k$ that is 1 for $0 \leq k \leq n$ and 0 for $n+1 \leq k \leq \infty$. Meh, whatever.
 Sci Advisor HW Helper Thanks P: 25,251 Can I call it 'finite power series'? What do you call it?
 Emeritus Sci Advisor PF Gold P: 5,534 I call it a sum. Edited to add: Actually, it exactly meets the definition of an n-th degree polynomial (with specific coefficients). So I'd call it a polynomial.
 P: 20 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. ) = $$\sum^n _{k=0} (3k - 3^k)$$ $$= 3 \sum^n _{k=0} k - \sum^n _{k=0} 3^k$$ $$= 3 (n(n + 1)/2) - 3n$$ $$= 3 - 3n$$ $$= n$$ for b.) $$\sum^{\infty} _{k=0} \frac{k-1}{2^k}$$ i have no answer yet... pls help. for c.) $$\sum^n _{k=1} \frac{-1}{k(k + 1)}$$ $$= \sum^n _{k=1} (\frac{k-1} {k(k + 1)}) - (\frac {1}{(k + 1)})$$ $$= 1 - 1 / n$$
 Sci Advisor HW Helper Thanks P: 25,251 The first term in a) is the only part of this that makes any sense. To make any further progress you are going to have to learn how to sum a geometric series $$\sum^n _{k=0} (x^k)$$. Could you look that up and post the answer and then apply it to a) to get a correct solution? Once you do that, this thread has all of the hints you need to do b). I think c) may be harder - but let's get a) and b) out of the way anyway. Can you also double check which (if any) of the limits is infinite? I rather hope it's c). And is the (-1) in c) supposed to be (-1)^k?
P: 20
 Quote by Dick The first term in a) is the only part of this that makes any sense. To make any further progress you are going to have to learn how to sum a geometric series $$\sum^n _{k=0} (x^k)$$. Could you look that up and post the answer and then apply it to a) to get a correct solution? Once you do that, this thread has all of the hints you need to do b). I think c) may be harder - but let's get a) and b) out of the way anyway. Can you also double check which (if any) of the limits is infinite? I rather hope it's c). And is the (-1) in c) supposed to be (-1)^k?
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?

$$= 3 \sum^n _{k=0} k - \sum^n _{k=0} 3^k$$

then what should i do after this? Kindly post example problems and further solutions to explain it well, please.

thanks a lot.
HW Helper
Thanks
P: 25,251
 Quote by nightcrrawlerr 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? $$= 3 \sum^n _{k=0} k - \sum^n _{k=0} 3^k$$
No, it's not a telescoping series, but your smplification is fine. You know what to do with the first term, right? Use the same formula you used before. As I keep saying, there is also a formula for the second term. It's called a 'geometric series'. Find out about it!
P: 20
 Quote by Dick No, it's not a telescoping series, but your smplification is fine. You know what to do with the first term, right? Use the same formula you used before. As I keep saying, there is also a formula for the second term. It's called a 'geometric series'. Find out about it!
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)

$$(1 - a)\sum^n _{k=0} 3^k = \sum^n _{k=0} 3^k - \sum^n _{k=0} 3^k+^1$$

$$= \sum^n _{k=0} 3^k - \sum^n+^1 _{k=1} 3^k$$

$$= 3 + 3^1 + 3^2 + ... + 3^n$$
$$- 3^1 - 3^2 - ... - 3^n - 3^n+1$$

$$= 3 - 3^n+1$$
$$= 3 - 3^n+^1$$

$$(1 - a) \sum^n _{k=0} 3^k = 3 - 3^n+^1/ 1 - a$$

$$\sum^n _{k=0} 3^k = 3 - 3^n+^1 / 1 - a$$

for P "not equal" 1
P: 20
 Quote by Dick No, it's not a telescoping series, but your smplification is fine. You know what to do with the first term, right? Use the same formula you used before. As I keep saying, there is also a formula for the second term. It's called a 'geometric series'. Find out about it!
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)

$$(1 - a)\sum^n _{k=0} 3^k = \sum^n _{k=0} 3^k - \sum^n _{k=0} 3^k^+^1$$

$$= \sum^n _{k=0} 3^k - \sum ^{^n^+^1} _{k=1} 3^k$$

$$= 3 + 3^1 + 3^2 + ... + 3^n$$

$$= - 3^1 - 3^2 - ... - 3^n - 3^n^+^1$$

$$= 3 - 3^n^+^1$$

$$(1 - 3) \sum^n _{k=0} 3^k = 3 - 3^n^+^1/ 1 - 3$$

$$\sum^n _{k=0} 3^k = 3 - 3^n^+^1 / 1 - 3$$

for a "not equal" 1

kindly please comment on this pls.
 Sci Advisor HW Helper Thanks P: 25,251 That looks good. Now that you've got part a) look at part b). There's a pretty strong hint back in the thread. And as Tom suggested so long ago, I think c) is best treated as telescoping series.
P: 20
 Quote by Dick That looks good. Now that you've got part a) look at part b). There's a pretty strong hint back in the thread. And as Tom suggested so long ago, I think c) is best treated as telescoping series.
So you mean got the solution correct? how about the first term, how can i combine the answers in one equation?

thank you.
P: 20
 Quote by Dick The first term in a) is the only part of this that makes any sense. To make any further progress you are going to have to learn how to sum a geometric series $$\sum^n _{k=0} (x^k)$$. Could you look that up and post the answer and then apply it to a) to get a correct solution? Once you do that, this thread has all of the hints you need to do b). I think c) may be harder - but let's get a) and b) out of the way anyway. Can you also double check which (if any) of the limits is infinite? I rather hope it's c). And is the (-1) in c) supposed to be (-1)^k?

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.
 Mentor P: 40,658 Hey, do y'all want me to move this to a General Math forum? I probably made a mistake putting it here. If it would help at all for this to be in a General Math forum, please let me know. Thanks.
P: 20
 Quote by berkeman Hey, do y'all want me to move this to a General Math forum? I probably made a mistake putting it here. If it would help at all for this to be in a General Math forum, please let me know. Thanks.
Ok sure you will. No problem for me.

thanks
 Mentor P: 40,658 Thread moved to the general calculus forum. This is more general than the original impression that I had. Sorry about that. These are hard.
 P: 20 Guru, can i start posting my ans. on item c?
 P: 20 Dick can i post here my answer for c.)? c.) $$\sum^n _{k=1} \frac{-1}{k(k + 1)}$$ $$= \sum^n _{k=1} (\frac{-1}{k} + \frac {1}{k + 1})$$ $$= (\frac {-1}{1} + \frac {1}{2}) + (\frac {-1}{2} + \frac {1}{3}) + ... + (\frac{-1}{n} + \frac {1}{n + 1})$$ $$= {-1} + \frac{1}{n + 1}$$ cristo, am i doing it right? thanks for your help by the way.

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