# Summation of series

tyneoh

## Homework Statement

Let v1, v2, v3 be a sequence and let

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

for n= 1,2,3... find $\sumun$ from n=1 to N.

## The Attempt at a Solution

Began with method of differences and arrived at
Sn= v1-(n+1)vn+1

Mentor
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 + ...

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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 $\sumun$ from n=1 to N.

## The Attempt at a Solution

Began with method of differences and arrived at
Sn= v1-(n+1)vn+1

Karnage1993
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:
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...

## 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:
Mentor
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?

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