Summation of Series Homework: Find Sn

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Homework Help Overview

The problem involves a sequence defined by the expression un = nvn - (n+1)vn+1, and participants are tasked with finding the sum of this sequence from n=1 to N. The context is within the study of series and summation in mathematics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the method of differences and consider writing out the terms of the sequence to identify potential cancellations in the summation. There is also an exploration of deriving a general formula for the sum based on specific terms calculated.

Discussion Status

There are multiple attempts to articulate the sum of the series, with some participants confirming the results of others. Questions about the clarity of expressions and potential errors in notation have been raised, indicating an ongoing dialogue about the formulation of the problem.

Contextual Notes

Some participants note the recursive nature of the sequence and the need for clarity in the expressions used, suggesting that there may be confusion regarding the formulation of the sum.

tyneoh
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Homework Statement



Let v1, v2, v3 be a sequence and let

un=nvn-(n+1)vn+1

for n= 1,2,3... find \sumu<sub>n</sub> from n=1 to N.

Homework Equations




The Attempt at a Solution


Began with method of differences and arrived at
Sn= v1-(n+1)vn+1
 
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I'd first try writing out u1, u2, u3 to see if there's some term cancellations that come about when you sompute
sum (un) = u1 + u2 + u3 + ...
 
tyneoh said:

Homework Statement



Let v1, v2, v3 be a sequence and let

un=nvn-(n+1)vn+1

for n= 1,2,3... find \sumu<sub>n</sub> from n=1 to N.

Homework Equations




The Attempt at a Solution


Began with method of differences and arrived at
Sn= v1-(n+1)vn+1
What's your question?
 
I think the question is: given the recursive formula ##u_n = nv_n - (n+1)v_{n+1}##, find the general formula for the sum of ##u_1 + u_2 + ... + u_n## for any n.

Let's first list out some possibilities:

##u_1 = v_1 - 2v_2\\ u_2 = 2v_2 - 3v_3\\ u_3 = 3v_3 - 4v_4##

So the sum of the three is:

sum{##u_3##} ##= v_1 - 2v_2 + 2v_2 - 3v_3 + 3v_3 - 4v_4 = v_1 - 4v_4##

Based on this, can you think of a formula for any ##n##th sum?
 
Last edited:
tyneoh said:
...

The Attempt at a Solution


Began with method of differences and arrived at
Sn= v1-(n+1)vn+1
If you mean \displaystyle\ \ S_N=\sum_{n=1}^{N}u_n=v_1-(N+1)v_{N+1},,\ then your result looks good.
 
Last edited:
SammyS said:
If you mean \displaystyle\ \ S_N=\sum_{n=1}^{N}=v_1-(N+1)v_{N+1},,\ then your result looks good.

Do you have one too many equals?
 
jedishrfu said:
Do you have one too many equals?
LOL !

Thanks!

Actually I had one too few un .

I'll edit my post!
 

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