# Finding the sum of a series

1. Aug 11, 2013

### Miike012

I don't understand any of the steps that were made in the book.

So I will try and solve it on my own so please let me know where I am going wrong.

The problem is in the paint document.

The general Term for the infinite series given is n(n+1) where n is greater than one and integral.

LHS we have: which has (k-1) terms
n(n+1) + (n+1)(n+2) + (n+2)(n+3) +..... +(n+k)(n+k+1)

and (n+k)(n+k+1) = n2 + (2k+1)n + k2 + k (I will use the coefficient of n to find the upper value of the variable M below)

Expanding the LHS we and grouping like terms we have:

= (k-1)n2 + [Ʃ(1+2M)]n + constant: Note: the second term [Ʃ(1+2M)]n M ranges from 0 to k.

Now I have:
constant + [Ʃ(constant+2M)]n + (k-1)n2 = A + B(n+1) + C(n+1)2 +......

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2. Aug 11, 2013

### Miike012

Another question what happened to the terms Bn, Cn2, Dn3.... in the portion I highlighted in the paint document?

And the highest power on the LHS is 2 so why is the coefficient D not equal to zero?

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3. Aug 11, 2013

### junaid314159

They created the series with n terms and also the series with n+1 terms. The terms in both of these series are identical except for the last term. So when they subtract the two series, you are left with a nice simple LHS which is (n+1)(n+2). Originally A was the constant and B the coefficient on the linear term. After subtracting the A canceled out and the Bn canceled out, leaving B as the new constant.

The approach you are using is different as you are not subtracting two series to solve the problem. Therefore, in your approach the B is the coefficient on the linear term while it is the constant term in their approach. You are definitely right in looking at the degree of both sides and deciding which coefficients must be zero but keep in mind that it will be a little different in both approaches.

I also don't see how you went from (1+2M) to (constant+2M) in your summation.

-Junaid

4. Aug 11, 2013

### rcgldr

In the image, the RHS of that equation is the RHS of the second equation - the RHS of the first equation. As far as which coefficients on the RHS should be zero, you have to assume that the statement that A, B, C, D, E, ... are constant and independent of (n+k) is true.

5. Aug 11, 2013

### Miike012

Sorry I didn't mean to write that

6. Aug 11, 2013

### Miike012

I got it now. I was thinking that
1*2 + 2*3 + .... n(n+1) was equal to
1*2 + 2*3 + .... n(n+1) (n+1)(n+2)
because I was assuming that they were adding all the terms in the infinite series that was given therefore I thought they were just showing the next term in the infinite series. But in reality they are adding the first n terms of an infinite series and that's why they said " find the sum of the series 1*2 + 2*3 + .... n(n+1)." and not "find the sum of the series 1*2 + 2*3 + .... n(n+1) ....."

7. Aug 11, 2013

### rcgldr

In general, the sum of a polynomial series

Ʃ f(n) = Ʃ ck nk + ck-1 nk-1 + ... + c0

will be a polynomial of degree k+1:

Ʃ f(n) = g(n) = dk+1 nk+1 + dk nk + ... + d0

You could just create k+1 equations, perhaps for values 1 through k+1, and use linear algebra to solve for the d's.

The process you're doing for this problem is effectively proving this, by determining which coefficients should be zero.

Last edited: Aug 11, 2013
8. Aug 11, 2013

### junaid314159

Yes, that's an important point. You have to keep in mind that this problem is about finding the sum of a finite series, not an infinite series.

-Junaid

9. Aug 11, 2013

### rcgldr

After subtraction, D is the coefficent for (3n^2 + 3n + 1), so it's coefficients E, F, ..., that should be zero. So now you need to solve for A, B, C, and D.

Last edited: Aug 11, 2013
10. Aug 11, 2013

### Miike012

Thanks rcgldr and junaid314159 for your help. I now understand what's going on.

11. Aug 11, 2013

### Miike012

It's crazy that I've been through differential, integral, and multi var. calculus, dif equations and linear algebra and I find many of the problems in this algebra book ( Higher algebra by henry Sinclair) more difficult than many of the problems I've been introduced in any of my math classes I listed above.

12. Aug 11, 2013

### rcgldr

So were you able to solve for A, B, C, and D?

13. Aug 11, 2013

### Miike012

Yes it was easy after I understood what the book was talking about

14. Aug 11, 2013

### Miike012

The equations that I came up with were
4E = 0
6E + 3D = 1
4E + 3D + 2C = 3
B + C + D + E = 2

Then I solved For the unknowns then I went back to the original equation and solved for A

15. Aug 11, 2013

### rcgldr

For anyone else reading this, after subtraction

(n+1)(n+2) = B + C(2n+1) + D(3n^2 + 3n + 1) + E(4n^3 + ...) + ...
n^2 + 3n + 2 = B + C(2n+1) + D(3n^2 + 3n + 1) + E(4n^3 + ...) + ...

Since A, B, C, D, ... are independent of n, then E, F, ... = 0, since they involve terms of n^3 or greater power.

Then looking at the terms for n^2 in the above equation:

n^2 = 3 n^2 D
D = 1/3

Next the terms for n

3n = n (2C + 3D) = n (2C + 3(1/3)) = n (2C + 1) = 2n C + n
2n = 2n C
C = 1

Next the constant term

2 = B + C + D = B + 1 + 1/3
B = 2/3

Then going back to the original equation and using the constant term for the case where n = 1:

1*2 = 2 = A + B (1) + C (1) + D (1) = A + 2/3 + 1 + 1/3
A = 0

As mentioned in post #7, knowing that the series sum of some function of degree k will be a function of degree k+1, then the series can just be evaluated k+1 times to produce k+1 linear equations, which are then solved for the coefficients.

Code (Text):

1 a + n b + n^2 c + n^3 d = sum: i^2 + i ; for i = 1 to n

| 1  1  1  1 |  | a | = |  2 |
| 1  2  4  8 |  | b | = |  8 |
| 1  3  9 27 |  | c | = | 20 |
| 1  4 16 64 |  | d | = | 40 |

| a | = |     4    -6     4    -1 |  |  2 |
| b | = | -13/3  19/2    -7  11/6 |  |  8 |
| c | = |   3/2    -4   7/2    -1 |  | 20 |
| d | = |  -1/6   1/2  -1/2   1/6 |  | 40 |

| a | = |   0 |
| b | = | 2/3 |
| c | = |   1 |
| d | = | 1/3 |

Last edited: Aug 12, 2013