Sum of n Terms of 7/(1.2.3) - 17/(2.3.4) + Series

[IEVaibhov]

Find the sum of n terms of the series:

7/(1.2.3) - 17/(2.3.4) + 31/(3.4.5) - 49/(4.5.6) + 71/(5.6.7) - ...I know how problems like the following are solved :
1. 1/(1.2.3) + 2/(2.3.4) + 3/(3.4.5) + ...
2. 3/(1.2.4) + 4/(2.3.5) + 5/(3.4.6) + ...What will be the general term of the required series? How do I proceed exactly? I am not understanding how to start solving the problem.

 

[Ecthelion]

Series can definitely be tricky.

Things to look at first - with a series such you need to look at distinct patterns in the numbers given to you. The first thing that jumps out at my eye is that the sigh (+ or -) alternates each term. What way do series that do that are generally represented?

Next I would tackle this part:

\frac{1}{1*2*3} + \frac{1}{2*3*4} + \frac{1}{3*4*5} and see if you can't come up with a separate summation representation for that. At this point, you almost have all of it done.

The next part is to look at the integer portion at the top, so the [7, 17, 31, 49] part of the terms. See any relationships between them?

Hopefully that helps you get started.

 

[IEVaibhov]

If you consider the difference between the difference of the consecutive terms, it comes out to be constant.
7 17 31 49
10 14 18
4 4

Now what?

 

[Ecthelion]

Perfect. Now how would you incorporate that with getting those particular initial terms in a sum? How could you then make the sign change each time?

 

[IEVaibhov]

We need a (-1)^n term in the general term for the sign change, that is okay..
Now, I know 2 things :
1: 17 = 7+10
2: 31 = 17 + 14
3: 14 = 10 + 4
Let me put the above two equations in variable form.

Consider this -

7 = x₁
17 = x₂
31 = x₃


10 = y₁
14 = y₂

and

4 = z₁

Basically,

1 => x₂ = x₁ + y₁

2=> x₃ = x₂ + y₂

3=> y₂ = y₁ + z₁

Therefore, we have -

x₃ = x₁ + 2y₁ + z₁

 

[Bohrok]

If you consider the difference between the difference of the consecutive terms, it comes out to be constant.
7 17 31 49
10 14 18
4 4

Now what?

Don't look at the difference between the difference of the terms. If you just consider the difference of the terms, they form an arithmetic sequence: 10, 14, 18, 22, ...
Now take a look at this:

a₂ = a₁ + 10
a₃ = a₂ + 14 = a₁ + 10 + 14
a₄ = a₃ + 18 = a₁ + 10 + 14 + 18

It just becomes a₁ plus an arithmetic series.

Since a₁ = 7, try plugging everything into the formula S_n = n(a₁ + a_n)/2. You'll have to play around a bit with the formula you get, but then that'll take care of the numbers in the numerator.

 

[IEVaibhov]

To me, this looks like -
S_n = a₁ + 5*2 + 7*2 + ... (till n terms)
and a_n = a₁ + 2*(2n+1) ; n is a natural number that lies between 2 and n.

So, S_n = [n(2a₁ + 4n + 2)] / 2

(Since S_n = n(a₁ + a_n) / 2 )

Therefore, S_n = n[16 + 4n] /2
= 8n + 2n^2 ( n going from 2 to n)

Is that correct?

 

[vela]

If you plug in values for n, does your formula reproduce the sequence you're seeking?

 

[IEVaibhov]

Yes..

 

[vela]

Then it's correct.

 

[IEVaibhov]

Oh wait, I got the summation only for the numerator part.

 

[Ecthelion]

Well, see if you can take a stab at it.

The nature of the numbers should make you think factorials. Using that intuitive leap, try to logic out a way to - as vela suggested - plug in n and receive the correct values.

 

[Bohrok]

Oh wait, I got the summation only for the numerator part.

Don't forget the a₁ in the summation formula to get the numbers for the numerators. If b_n is the sequence of numbers for the numerators, b_n is basically going to be S_n once you've plugged in a₁. Then you just have the denominators to finish, and that's easy.