# Find the sum S(n)

1. Dec 17, 2004

### primarygun

How can I find the a function to express the sum of a function, like
$$1^2+2^2+3^2+....+n^2=?$$
What's the expression?
Anyway, how can I use latex?

Last edited: Dec 17, 2004
2. Dec 17, 2004

### Nylex

Isn't that sum just $$\sum_{n=1}^\infty n^2$$?

3. Dec 17, 2004

### matt grime

It's the sum from 1 to n of r^2, not an infinite sum, especially an infinite sum that obviously diverges.

Seach these forums for it as it comes up a lot.

4. Dec 17, 2004

### primarygun

With what keywords?

5. Dec 17, 2004

### CrankFan

6. Dec 17, 2004

### primarygun

Thanks. They are useful

7. Dec 17, 2004

### HallsofIvy

Staff Emeritus
For a problem like that, "Newton's finite Difference" method, a variation on Taylor's polynomial works nicely.

Make a list of values and repeated differences:
n y(n) dy d2y d3y d4y
0 0 1 3 2 0
1 1 4 5 2
2 5 9 7
3 14 16
4 30

Where the number in each "difference" column is the difference between two successive numbers in the previous column. Of course, since the y value itself is gotten by adding squares, the first difference is just the square itself. The second differences are just odd numbers and the third differences are the same: 2. All succeeding differences are 0.

Newton's polynomial now is y+ (dy)n+ (d2y/2) n(n-1)+ (d3y/3!)n(n-1)(n-2)+ ... with the ith term being (diy)/n! n(n-1)..(n-i+1)

Here, that gives 0+ 1n+ (3/2)n(n-1)+ (2/6)n(n-1)(n-2)= (n/6)(6+ 9n- 9+ 2n[sup[/sup]-6n+ 4)= (1/6)n(2n2+3n+1)= (1/6)n(2n+1)(n+1).

One can show that the sum of "kth" powers is a polynomial of degree k+1.

Last edited: Dec 17, 2004
8. Dec 17, 2004

### primarygun

I wish I have learnt calculus.

9. Dec 18, 2004

### CrankFan

One way to do it.

$$\sum_{k=1}^{n} k^2$$

$$\sum_{k=1}^{n} k^2 -k + k -1 + 1$$

$$\left( \sum_{k=1}^{n} k^2 - k - {1 \over 2} + 1 \right) + \left( \sum_{k=1}^{n} k - {1 \over 2} \right)$$

$$\left( \sum_{k=1}^{n} k^2 - k - {1 \over 2} + {1 \over 3 } + {2 \over 3} \right) + \left( {1 \over 2} \sum_{k=1}^{n} 2 \left(k - {1 \over 2} \right) \right)$$

$$\left( \sum_{k=1}^{n} k^2 - k + {1 \over 3} \right) + \left( \sum_{k=1}^{n} {2 \over 3} - {1 \over 2} \right) + \left( {1 \over 2} \sum_{k=1}^{n} 2k - 1 \right)$$

$$\left( {1 \over 3} \sum_{k=1}^{n} 3k^2 - 3k + 1 \right) + \left( \sum_{k=1}^{n} {1 \over 6} \right) + \left( {1 \over 2} \sum_{k=1}^{n} 2k - 1 \right)$$

$$\left( {1 \over 3} \sum_{k=1}^{n} k^3 - (k-1)^3 \right) + \left( \sum_{k=1}^{n} {1 \over 6} \right) + \left( {1 \over 2} \sum_{k=1}^{n} 2k - 1 \right)$$

$${k^3 \over 3} + {n \over 6} + {n^2 \over 2}$$

If you're wondering how $$\sum_{k=1}^{n} 2k - 1 = n^2$$

$$\sum_{k=1}^{n} 2k - 1$$

$$\sum_{k=1}^{n} k^2 - k^2 + 2k - 1$$

$$\sum_{k=1}^{n} k^2 - (k^2 - 2k + 1)$$

$$\sum_{k=1}^{n} k^2 - (k-1)^2$$

$$k^2$$

This last step is called telescoping property of sums, it was also
used in the last step of the first proof, for the leftmost term.

$$\sum_{k=1}^{n} (a_k - a_{k-1}) = a_n - a_0$$

Last edited: Dec 18, 2004
10. Dec 18, 2004

### primarygun

After reading the web site last night, I got it.

11. Dec 18, 2004

### tongos

1^2+2^2+3^2+4^2+5^2.......x^2

all you have to do is look at increasing three digit numbers, and the amount of combinations in an increasing three digit number is 1+2+3+4+5+6+7+(1+2+3+4+5+6)+(1+2+3+4+5).......

but there is any easier way to express this, it could be 9(8)(7)/6 because an increasing number is only one out of six.

now look 100(1)+/99(2)/+98(3)+/97(4)/+96(5)+/95(6)/.........+/1(100)/= 102(101)(100)/6

and 99(1)+97(2)+95(3)+93(5)...............+1(50)

so the terms in / / are going to be repeated again, and you divide that by two so you get the series below, and you get (102)(101)(100)/24 as the sum which is (2n+2)(2n+1)(2n)/24 which is equal to the sum of the nth squares.
kind of a cool way. to find the sum of cubes, use four digit numbers.