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38510987654321 222222222210
1
2 n
This total represents the number of blocks required to build a pyramid of ten layers.
38510987654321 222222222210
1
2 n
But doing so teaches you nothing about how power series behave in general. There is a theoretical approach that I like and which I have derived from first principles using a ton of algebra. I am going to share it over the next few pages. Be warned that there is a LOT of algebra involved, in fact some of the longest equations I have seen in years.
This could easily be worked out in a matter of minutes using a spreadsheet. A spreadsheet is definitely the fastest method for adding up a power series, if you take into consideration the time needed to go fetch a more theoretical method off the internet.
2222
1
2 ...321 NnN
Start by considering the graph of the totals as the number of terms increases. It looks as though the sum of a power series is another power series.
DCNBNANnN
23
1
2
2222
1
2 ...321 NnN
This is borne out by looking at the differences between terms, and then the differences between differences. If this eventually reaches a line of unchanging numbers, then the series is a polynomial function.
In this case, the constant line appears at the third difference, which means that the polynomial is a cubic.
DCNBNANnN
23
1
2
The other terms will be harder to work out, however.
Common sense tells us that D in this function has to be zero. The sum of no bricks has to be none!
14333
5222
1111
23
23
23
CBA
CBA
CBA
CNBNANnN
23
1
2One way to do it would be to look at any three (x,y) pairs and set up a 3x4 matrix.
This could definitely be done, but it would be a horrible process.
14333
5222
1111
23
23
23
CBA
CBA
CBA
CNBNANnN
23
1
2
A better way is to compare the (N+1)th term with the Nth.
21
21
1
2 1
NnnNN
111 1 23232CNBNANNCNBNAN
1
1
1 1
22
332
NNC
NNB
NNAN
1
12
133 12 22
C
NB
NNANN
Compare coefficients:
1
12
133 12 22
C
NB
NNANN
1
223
13
CBA
BA
A
This might look messy but it is set up so that it unzips nicely. Each coefficient can be identified directly, one line at a time.
61
21
31
C
B
A
NNNnN
612
213
31
1
2
This unzipping approach can be used to reveal the coefficients on higher power series. For example, the sum of cubic terms:
31
31
1
3 1
NnnNN
1 1 1 1 1 2233443NNDNNCNNBNNAN
DNCNBNANnN
234
1
3
0.1 1.1 0.0.1 .2.11 0.0.0.1 .3.3.11 1.0.0.0.0 - .4.6.4.11 1.3.3.1 DCBA
In this notation, only the coefficients are expressed. The rest is implied by the context.
1 1 1 1 1 2233443NNDNNCNNBNNAN
Expanding the bracketed terms in the normal way will turn this into an equation so big it won’t even fit on the page! But I can avoid this by introducing a better notation.
You can see that the end result is the combinatorial coefficients less the first term, obtained (for example) from Pascal’s triangle.
1 12. 3.3.1 4.6.4.1 1.3.3.1 DCBA
0.1 1.1 0.0.1 .2.11 0.0.0.1 .3.3.11 1.0.0.0.0 - .4.6.4.11 1.3.3.1 DCBA
If structured in a matrix, things get even clearer.
DCBA
CBAN
BAN
AN
1 1 1 1 1 : 1
2 3 4 3 :
3 6 3 :
4 1 :
2
3
The coefficients unzip from the top line downwards.
2
413
214
41
1
3 NNNnN
1111
234
36
4
1
3
3
1
D
C
B
A
1 12. 3.3.1 4.6.4.1 1.3.3.1 DCBA
NNNNNnN
n
30123
314
215
51
1
4 0
EDCBA
DCBAN
CBAN
BAN
AN
111111:1
23454:
36106:
4104:
51:
2
3
4
The same idea should apply to the next series up, namely the sum of all quartics:
... and the sum of quintics.
No, I don't want to solve that set of equations by hand either! But a spreadsheet makes quick work of it.
FEDCBA
EDCBAN
DCBAN
CBAN
BAN
AN
1111111:1
234565:
36101510:
4102010:
5155:
61:
2
3
4
5
NNNNNNnN
n
00 2
12134
1255
216
61
1
5
... and so on.
Some things to notice about all of this.
Fact 1: The leading coefficient is always the integral coefficient.
1
1
2
2
1
11
1
... NANANANn P
P
P
P
NP
This doesn’t surprise me in the least. As N gets very large, the summation starts to resemble an integration. The term with the highest exponent pulls away from the lesser terms. Therefore, in the extreme case where N tends towards infinity, the leading coefficient has to equal the integral coefficient.
Fact 2: The second coefficient is always ½. This really IS a surprise.
Some things to notice about all of this.
1
1
2
2211
11
1
... NANANNn PP
P
NP
Fact 3: The coefficients always add to 1.
If you can remember those three facts, then it will be easy to recall the coefficients for the first two power sums.
1
212
21
1
1 NNnN
1
612
213
31
1
2 NNNnN
And the good news is that these are the summations you are most likely to come across. We rarely if ever see the sums of cubics or quartics or higher powers.
There is a relationship between the sum for cubics and the sum for linear terms.
2
1
2
2
2
413
214
41
1
3 1
N
N
N
nNNNNn
This shortcut gives you one more summation that’s easy to recall.
It’s kind of hard to imagine what the sum of cubic terms could be applied to. The sum of linear terms is the number of logs stacked in a triangle. The sum of square terms is the number of bricks stacked in a pyramid (a three-dimensional triangle?). Therefore the sum of cubic terms is the number of hyperbricks stacked in a four-dimensional triangle, a hyper-pyramid.
That’s all.
- David C, 2015
