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### Finding Sum Formula using Sequences of Differences

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Date: 06/28/98 at 20:05:12
From: Kyungsoo
Subject: Sum equation for higher powers

Hello!

What is the SUM FORMULA for

infinity (n)
sum(sigma)   k^4
k=1

For example, the sum formula for

infinity(n)
sum(sigma)   k     is n(n+1)/2
k=1

The sum formulas I know are only up to k^3.

Then what are the sum formulas for higher powers of k?

Thanks,

KyungSoo
```

```
Date: 07/04/98 at 14:11:45
From: Doctor Floor
Subject: Re: Sum equation for higher powers

Hi KyungSoo,

Thank you for sending your question to Dr. Math. An interesting
question it is!

I will show you how to find the formula for

n
SUM k^4
k=1

so you can do it for higher powers yourself. By the way, I'd rather
not use the word infinity here, because you want n to be finite. If n
were infinite, the sum would be infinite, too.

To make it a little easier for ourselves, I prefer to make the formula
for
n
S_n =  SUM k^4
k=0

which doesn't make any difference at all, except that it starts with
k=0 and is defined for n=0.

The formula that we will find for S_n will be a polynomial. The first
thing we have to do is find out the degree of this polynomial. You can
find the degree of this polynomial sequence by making so-called
"sequences of differences." You make a sequence of differences by
subtracting the consecutive numbers of the original sequence. As an
example, I have taken the values of n^2 for n=1 to n=5. You'd begin by
writing out these terms into the first line and then subtracting
consecutive terms to write the second line.

1  4  9  16  25
3  5  7   9    <-- (first) sequence of differences.
^
|
3 = 4-1

Now if you make a sequence of differences (SoD) of a polynomial
sequence, then the degree of this SoD is one lower. In my example, you
can see that the SoD is linear, so it is of first degree, whereas the
original sequence was of second degree.

To see that this holds for the general case, suppose we have a
sequence defined by P(n) = a*n^t + Q(n), where Q(n) is of lower degree
than P(n). Then P(n+1) = a*(n+1)^t + Q(n+1) = a*n^t + R(n) + Q(n+1),
where R(n) is the remainder of a*(n+1)^t when you have taken a*n^t
out. R(n) is of degree t-1.

But for the sequence of differences (SoD(n)) of P(n) this means:
SoD(n) = P(n+1) - P(n) =
= a*n^t + R(n) + Q(n+1) - (a*n^t + Q(n))
= R(n) + Q(n+1) - Q(n).

So the degree of SoD(n) is lower than the degree of P(n), because R(n)
and Q(n) (and Q(n+1)) are of lower degree than P(n). But we have not
yet established that the degree is exactly one lower. It is exactly
one lower, because Q(n+1)-Q(n) makes the SoD of Q(n), and thus Q(n+1)-
Q(n), of lower degree than Q(n)! So Q(n+1)-Q(n) is of degree t-2 or
less. And since R(n) is of degree t-1, this must also be the case for
SoD(n).

Back to the sequence S_n we are working on. This must be of fifth
degree now, since S_n - S_(n-1) = n^4, so that the first SoD is
clearly of fourth degree. You can also see that if you make
consecutive SoD's - SoD's of SoD's, that is - then the fifth SoD is
constant (degree zero). This is another way to see that our sequence
must be polynomial of fifth degree. So let's calculate our sequence of
sums of fourth power and check this out.

S_0 = 0^4 = 0
S_1 = S_0 + 1^4 = 0 + 1 = 1
S_2 = S_1 + 2^4 = 1 + 16 = 17
S_3 = S_2 + 3^4 = 17 + 81 = 98
S_4 = S_3 + 4^4 = 98 + 256 = 354
S_5 = S_4 + 5^4 = 354 + 625 = 979
S_6 = S_5 + 6^4 = 979 + 1296 = 2275

And now put these into the first line, and make five SoD's!

0    1    17    98    354    979    2275 <-- S_n
1   16    81    256    625    1296
15   65    175   369    671
50    110   194    302
60    84    108
24    24      <-- 5th SoD is constant

Now that we know that the formula will be a fifth degree formula, it
has to be of the form

S(n)= an^5 + bn^4 + cn^3 + dn^2 + en + f.

Since S(0)=0 (here is where I made it a little easier!), you know that
f=0.

You could do two things now. The first is: make five equations with a,
b, c, d, and e by taking different values for n. You could solve this,
but it is a lot of work and no fun at all. The second thing you could
do is use that S(n) - S(n-1) = n^4. That would mean that you'd have to
expand S(n-1), which is not really fun, and then again you can make
five equations from S(n) - S(n-1) = n^4 (from the fact that the
coefficient with n^4 is 1, with n^3 is 0, etc.). I think these
equations will work faster, though. So I choose this second option.

First expand
S(n-1) = a(n-1)^5 + b(n-1)^4 + c(n-1)^3 + d(n-1)^2 + e(n-1).

Now we know that

(n-1)^2 = n^2 - 2n + 1
(n-1)^3 = n^3 - 3n^2 + 3n - 1
(n-1)^4 = n^4 - 4n^3 + 6n^2 - 4n + 1
(n-1)^5 = n^5 - 5n^4 + 10n^3 - 10 n^2 + 5n - 1

The coefficients come from Pascal's triangle.

This gives

S(n-1) = an^5 + (-5a+b)n^4 + (10a-4b+c)n^3 + (-10a+6b-3c+d)n^2
+ (5a-4b+3c-2d+e)n + (-a+b-c+d-e).

So

S(n) - S(n-1) = (5a)n^4 + (-10a+4b)n^3 + (10a-6b+3c)n^2
+ (-5a+4b-3c+2d)n + (a-b+c-d+e).

Since this has to equal n^4, you get the following equations:

5a = 1
-10a + 4b = 0
10a - 6b + 3c = 0
-5a + 4b - 3c + 2d = 0
a - b + c - d + e = 0

Solving these is quickly done, when just going down one after one:
a = 1/5, b = 1/2, c = 1/3, d = 0, e = -1/30.

So that would give

S(n) = (1/5)n^5 + (1/2)n^4 + (1/3)n^3 - (1/30)n.

I'll leave it for you to check this one, and to make formulas for
higher powers in this way.

Thank you for letting me experience this solution - I did not know
beforehand where it would end, and I was delighted by the simple
result. I hope you like the answer, too.

If you have other math questions, please send them to Dr. Math!

Best regards,

- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Polynomials
High School Sequences, Series

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