Summation of Series Homework: Find Sn

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The discussion focuses on finding the sum of the series defined by the recursive formula un = nvn - (n+1)vn+1 for n = 1, 2, 3, ... The initial attempts involved using the method of differences, leading to the expression Sn = v1 - (N+1)vn+1 for the sum from n=1 to N. Participants explored specific cases for u1, u2, and u3 to identify potential term cancellations. The conversation confirmed the correctness of the derived formula for the sum, with some clarification on the notation used. Overall, the goal is to establish a general formula for the sum of the series based on the recursive definition.
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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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