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

In summary, Find the sum of n terms of a series where the terms alternate (+ or -) each term. First, look for distinct patterns in the numbers given to you. Next, tackle the 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. Lastly, 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.
  • #1
IEVaibhov
15
0
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.
 
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  • #2
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:

[tex]\frac{1}{1*2*3} + \frac{1}{2*3*4} + \frac{1}{3*4*5} [/tex] 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.
 
  • #3
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?
 
  • #4
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?
 
  • #5
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
 
  • #6
IEVaibhov said:
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 Sn = n(a1 + an)/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.
 
  • #7
To me, this looks like -
Sn = a1 + 5*2 + 7*2 + ... (till n terms)
and an = a1 + 2*(2n+1) ; n is a natural number that lies between 2 and n.

So, Sn = [n(2a1 + 4n + 2)] / 2

(Since Sn = n(a1 + an) / 2 )

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

Is that correct?
 
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  • #8
If you plug in values for n, does your formula reproduce the sequence you're seeking?
 
  • #9
Yes..
 
  • #10
Then it's correct.
 
  • #11
Oh wait, I got the summation only for the numerator part.
 
  • #12
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.
 
  • #13
IEVaibhov said:
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 bn is the sequence of numbers for the numerators, bn is basically going to be Sn once you've plugged in a1. Then you just have the denominators to finish, and that's easy. :smile:
 

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

1. What is the formula for finding the sum of n terms of the given series?

The formula for finding the sum of n terms of the given series is Sn = (7/6) - (17/(2n+1)!).

2. How do you simplify the given series to find the sum of n terms?

To simplify the given series, we can use the formula for the sum of n terms, which is Sn = (a1(1-r^n))/(1-r), where a1 is the first term and r is the common ratio. In this case, a1 = 7/6 and r = -1/6. Plugging these values into the formula gives us Sn = (7/6(1-(-1/6)^n))/(1-(-1/6)). This can then be simplified further to Sn = (7/6(1-(-1)^n))/(7/6), which simplifies to Sn = 1-(-1)^n.

3. Is there a general pattern for finding the sum of n terms of this series?

Yes, there is a general pattern for finding the sum of n terms of this series. The general pattern is 1-(-1)^n, where n is the number of terms in the series. This pattern can be derived from the formula for the sum of n terms.

4. Can the sum of n terms of this series be negative?

Yes, the sum of n terms of this series can be negative. This is because the terms in the series alternate between positive and negative values, and the sum depends on the number of terms being added. If the number of terms is odd, the sum will be negative, and if the number of terms is even, the sum will be positive.

5. Can this series be used to find the value of pi?

No, this series cannot be used to find the value of pi. The series involves the terms 7 and 17, which are not related to the value of pi. To find the value of pi, we would need to use a different series or formula, such as the Leibniz formula or the Basel problem formula.

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