Method of differences for a series

[Eveflutter]

When using the method of differences on a given series, when do you stop listing the terms?

Example question:
f(r)=

; r∈N
State f(r)-f(r+1) in terms of r and hence determine

So skipping until the worked answer gives

Great so here I included the n+1^th term because I'm guessing since the larger number in the given: f(r)-f(r+1) was r+1 so that's why I stop here? I'm not sure.

Okay so that guess works for that one but what about this one:

Example question (2):
Given f(r)= r(r+1)! show that f(r)-f(r-1)= r!(r² +1) and hence solve

Alright skipping the proof once again and going into the expansion part, I have this thus far:
(3!2 - 2!1) + (4!3 - 3!2) + (5!4 + 4!3) + ... + (here is where I'm not sure what to stop at)

Possibilities:
stopping with the subbing of the n_th term ... + ((n-1)n! - (n-1)!(n-2)) + (n(n+1)! - n!(n-1))
or stopping with the subbing of the (n+1)_th term ... + ((n+1)!n - n!(n-1)) + ((n+2)(n+1)! - (n+1)!n)

The respective answers I may end up with are:
at n^th term: n(n+1)! - 2
at (n+1)^th term: (n+1)(n+1)! - 2

Any help will be appreciated ^3^ Thank you~!

 

[haruspex]

f(r)-f(r-1)= r!(r² +1)

For the purposes of the summation, the first term you want on the right is with r=2, so that must be the first term to use on the left:
f(2)-f(1).
One of those will cancel with something in the corresponding expression for the second term of the sum. Which one does not cancel?
Similarly for the last term in the sum.

 

[Eveflutter]

For the purposes of the summation, the first term you want on the right is with r=2, so that must be the first term to use on the left:
f(2)-f(1).
One of those will cancel with something in the corresponding expression for the second term of the sum. Which one does not cancel?
Similarly for the last term in the sum.

Alright...so I'm not completely following. The f(1) does not cancel at the beginning. But what I'm asking is: If I use either r=n or r=n+1 at the end there will be something different that does not cancel either way. That is the f(n) and the f(n+1) term respectively. Which term to stop at when listing them?

 

[haruspex]

Alright...so I'm not completely following. The f(1) does not cancel at the beginning. But what I'm asking is: If I use either r=n or r=n+1 at the end there will be something different that does not cancel either way. That is the f(n) and the f(n+1) term respectively. Which term to stop at when listing them?

Write out the last term in the sum on the right.
Write out f(r)-f(r+1) (in the first problem you posted).
What value of r makes these equal (up to a constant factor)?

 

[Eveflutter]

r must be n for both side to be equal...

 

[haruspex]

r must be n for both side to be equal...

Right. And of the f(r) pairs, which is the uncancelled member of the pair at r=n?

 

[Eveflutter]

n(n+1)! at the end?

 

[haruspex]

n(n+1)! at the end?

Actually I specified the first problem, but no matter.
Yes. The last term in the sum is (n²+1)n!, and this corresponds to f(n)-f(n-1). When you sum the f() terms the f(n) term there is the one that does not get cancelled.
Similarly, at the start of the sum we have f(2)-f(1)=(2²+1)2!, and the uncancelled f there is -f(1). Hence the sum is f(n)-f(1).

 

[Eveflutter]

Oh sorry ^3^" Got ahead of myself there. Alright so to sum it up, we stop when we reach the n term

When using the method of differences on a given series, when do you stop listing the terms?

Example question:
f(r)= View attachment 223786; r∈N
State f(r)-f(r+1) in terms of r and hence determine View attachment 223787

So skipping until the worked answer gives View attachment 223788
Great so here I included the n+1^th term because I'm guessing since the larger number in the given: f(r)-f(r+1) was r+1 so that's why I stop here? I'm not sure.

Okay so that guess works for that one but what about this one:

Example question (2):
Given f(r)= r(r+1)! show that f(r)-f(r-1)= r!(r² +1) and hence solve View attachment 223789

Alright skipping the proof once again and going into the expansion part, I have this thus far:
(3!2 - 2!1) + (4!3 - 3!2) + (5!4 + 4!3) + ... + (here is where I'm not sure what to stop at)

Possibilities:
stopping with the subbing of the n_th term ... + ((n-1)n! - (n-1)!(n-2)) + (n(n+1)! - n!(n-1))
or stopping with the subbing of the (n+1)_th term ... + ((n+1)!n - n!(n-1)) + ((n+2)(n+1)! - (n+1)!n)

The respective answers I may end up with are:
at n^th term: n(n+1)! - 2
at (n+1)^th term: (n+2***)(n+1)! - 2

Any help will be appreciated ^3^ Thank you~!

*** had a mistake here

 

[haruspex]

so to sum it up, we stop when we reach the n term

In this instance, yes.

 

[Eveflutter]

but to make sure check what value r should be for both sides to cancel like we did first?

 

[haruspex]

but to make sure check what value r should be for both sides to cancel like we did first?

Yes. In each case, you need to go through the steps I took.

 

[Eveflutter]

Okay great! Thank you so much for the help! I really appreciate it :D