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Simplifying summations 
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#19
Feb107, 12:15 PM

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RHS=Right Hand Side LHS=Left Hand Side 


#20
Feb107, 12:19 PM

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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 [itex]a_k[/itex] that is 1 for [itex]0 \leq k \leq n[/itex] and 0 for [itex]n+1 \leq k \leq \infty[/itex]. Meh, whatever. 


#21
Feb107, 12:22 PM

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Can I call it 'finite power series'? What do you call it?



#22
Feb107, 12:23 PM

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I call it a sum.
Edited to add: Actually, it exactly meets the definition of an nth degree polynomial (with specific coefficients). So I'd call it a polynomial. 


#23
Feb207, 05:06 AM

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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. ) = [tex] \sum^n _{k=0} (3k  3^k) [/tex] [tex] = 3 \sum^n _{k=0} k  \sum^n _{k=0} 3^k [/tex] [tex] = 3 (n(n + 1)/2)  3n [/tex] [tex] = 3  3n [/tex] [tex] = n [/tex] for b.) [tex] \sum^{\infty} _{k=0} \frac{k1}{2^k} [/tex] i have no answer yet... pls help. for c.) [tex] \sum^n _{k=1} \frac{1}{k(k + 1)} [/tex] [tex] = \sum^n _{k=1} (\frac{k1} {k(k + 1)})  (\frac {1}{(k + 1)}) [/tex] [tex] = 1  1 / n [/tex] 


#24
Feb207, 07:56 AM

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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 [tex] \sum^n _{k=0} (x^k) [/tex]. 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?



#25
Feb207, 11:40 PM

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Ok i'll try first your comment on this and i will post it later on. Lets start first at letter a) [tex] \sum^n _{k=0} (3k  3^k) [/tex] this can be solved by telescoping series right? does this simplifying make sense sir? [tex] = 3 \sum^n _{k=0} k  \sum^n _{k=0} 3^k [/tex] then what should i do after this? Kindly post example problems and further solutions to explain it well, please. thanks a lot. 


#26
Feb307, 10:12 AM

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#27
Feb507, 03:28 AM

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[tex] = \sum^n _{k=0} a^k = a + a^1 + a^2 + ... + a^n [/tex] for a "not equal" 1 [tex] = \sum^n _{k=0} 3^k = 3 + 3^1 + 3^2 + ... + 3^n [/tex] multiply both sides with (1a) [tex](1  a)\sum^n _{k=0} 3^k = \sum^n _{k=0} 3^k  \sum^n _{k=0} 3^k+^1[/tex] [tex]= \sum^n _{k=0} 3^k  \sum^n+^1 _{k=1} 3^k [/tex] [tex]= 3 + 3^1 + 3^2 + ... + 3^n[/tex] [tex]  3^1  3^2  ...  3^n  3^n+1[/tex] [tex]= 3  3^n+1[/tex] [tex]= 3  3^n+^1[/tex] [tex](1  a) \sum^n _{k=0} 3^k = 3  3^n+^1/ 1  a[/tex] [tex]\sum^n _{k=0} 3^k = 3  3^n+^1 / 1  a[/tex] for P "not equal" 1 


#28
Feb507, 03:30 AM

P: 20

[tex] = \sum^n _{k=0} a^k = a + a^1 + a^2 + ... + a^n [/tex] for a "not equal" 1 [tex] = \sum^n _{k=0} 3^k = 3 + 3^1 + 3^2 + ... + 3^n [/tex] multiply both sides with (1a) [tex](1  a)\sum^n _{k=0} 3^k = \sum^n _{k=0} 3^k  \sum^n _{k=0} 3^k^+^1[/tex] [tex]= \sum^n _{k=0} 3^k  \sum ^{^n^+^1} _{k=1} 3^k [/tex] [tex]= 3 + 3^1 + 3^2 + ... + 3^n[/tex] [tex]=  3^1  3^2  ...  3^n  3^n^+^1[/tex] [tex]= 3  3^n^+^1[/tex] [tex](1  3) \sum^n _{k=0} 3^k = 3  3^n^+^1/ 1  3[/tex] [tex]\sum^n _{k=0} 3^k = 3  3^n^+^1 / 1  3[/tex] for a "not equal" 1 kindly please comment on this pls. 


#29
Feb507, 08:23 AM

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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.



#30
Feb507, 07:15 PM

P: 20

Kindly help me more, please. thank you. 


#31
Feb507, 09:31 PM

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


#32
Feb607, 12:30 AM

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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.



#33
Feb607, 12:34 AM

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thanks 


#34
Feb607, 12:38 AM

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Thread moved to the general calculus forum. This is more general than the original impression that I had. Sorry about that. These are hard.



#35
Feb607, 04:05 AM

P: 20

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



#36
Feb607, 04:13 AM

P: 20

Dick can i post here my answer for c.)?
c.) [tex] \sum^n _{k=1} \frac{1}{k(k + 1)} [/tex] [tex]= \sum^n _{k=1} (\frac{1}{k} + \frac {1}{k + 1}) [/tex] [tex]= (\frac {1}{1} + \frac {1}{2}) + (\frac {1}{2} + \frac {1}{3}) + ... + (\frac{1}{n} + \frac {1}{n + 1}) [/tex] [tex]= {1} + \frac{1}{n + 1} [/tex] cristo, am i doing it right? thanks for your help by the way. 


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