Finding the sum of un = nvn - (n+1)vn+1 using telescoping

[tyneoh]

Homework Statement

Let v₁, v₂, v₃ be a sequence and let

u_n=nv_n-(n+1)v_n+1

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

Homework Equations

The Attempt at a Solution

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

 

[jedishrfu]

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 + ...

 

[SammyS]

Homework Statement

Let v₁, v₂, v₃ be a sequence and let

u_n=nv_n-(n+1)v_n+1

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

Homework Equations

The Attempt at a Solution

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

What's your question?

 

[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?

 

[SammyS]

...

The Attempt at a Solution

Began with method of differences and arrived at
S_n= v₁-(n+1)v_n+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.

 

[jedishrfu]

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?

 

[SammyS]

Do you have one too many equals?

LOL !

Thanks!

Actually I had one too few u_n .

I'll edit my post!