Because all the numbers in Pascal's Triangle are made the same way - by adding the two numbers
above and to the left
on the
row above
, then we can see that each red number is just the sum of a green number and a blue number and we use up all the
blue and green numbers to make all the red ones.
The sum of all the red numbers is therefore the same as the sum of all the blues and all the greens: 5+8=13!
The general principle that we have just illustrated is:
The
sum
of the numbers on one diagonal is the sum of the numbers on the previous two diagonals.
If we let D(i) stand for the sum of the numbers on the Diagonal that starts with one of the extra zeros at the beginning of row i,
then
D(0)=0 and D(1)=1
are the two initial diagonals shown in the table above. The green diagonal sum is D(5)=5 (since its extra initial zero is in row
5) and the blue diagonal sum is D(6) which is 8. Our red diagonal is D(7) = 13 = D(6)+D(5).
We also have shown that this is always true: one diagonals sum id the sum of the previous two diagonal sums, or, in terms of
our D series of numbers:
D(i) = D(i-1) + D(i-2)
But...
D(0) = 1
D(1) = 1
D(i) = D(i-1) + D(i-2)
is exactly the definition of the Fibonacci numbers! So D(i) is just F(i) and
the sums of the diagonals in Pascal's Triangle are the Fibonacci numbers!
Another arrangement of Pascal's Triangle
(8 of 25) [12/06/2001 17:13:40]

The mathematics of the Fibonacci series
By drawing Pascal's Triangle with all the rows moved over by 1 place, we have a clearer arrangement which shows the
Fibonacci numbers as sums of columns:
0
1
2
3
4
5
6
7
8
9
0
1
.
.
.
.
.
.
.
.
.
1
.
1
1
.
.
.
.
.
.
.
2
.
.
1
2
1
.
.
.
.
.
3
.
.
.
1
3
3
1
.
.
.
4
.
.
.
.
1
4
6
4
1
.
5
.
.
.
.
.
1
5 10 10
5
6
.
.
.
.
.
.
1
6 15 20
7
.
.
.
.
.
.
.
1
7 21
8
.
.
.
.
.
.
.
.
1
8
9
.
.
.
.
.
.
.
.
.
1
1
1
2
3
5
8 13 21 34 55
... <- sums of columns
This table can be explained by referring to one of the
(Easier) Fibonacci Puzzles
- the one about
Fibonacci for a Change
. It
asks how many ways you can pay n pence (in the UK) using only 1 pence and 2 pence coins. The order of the coins matters, so
that 1p+2p will pay for a 3p item and 2p+1p is counted as a different answer. [We now have a new
two pound coin
that is
increasing in circulation too!]
Here are the answers for paying up to 5p using only 1p and 2p coins:
1p
2p
3p
4p
5p
1p
2p
1p+1p
1p+2p
2p+1p
1p+1p+1p
2p+2p
1p+1p+2p
1p+2p+1p
2p+1p+1p
1p+1p+1p+1p
1p+2p+2p
2p+1p+2p
2p+2p+1p
1p+1p+1p+2p
1p+1p+2p+1p
1p+2p+1p+1p
2p+1p+1p+1p
1p+1p+1p+1p+1p
1 way
2 ways 3 ways
5 ways
8 ways
Let's look at this another way - arranging our answers according to
the number of 1p and 2p coins we use
. Columns will
represent all the ways of paying the amount at the head of the column, as before, but now the rows represent
the number of
coins in the solutions
:
cost:
1p 2p
3p
4p
5p
1 coin: 1p 2p
2 coins:
1p+1p 1p+2p
2p+1p
2p+2p
3 coins:
1p+1p+1p 1p+1p+2p
1p+2p+1p
2p+1p+1p
1p+2p+2p
2p+1p+2p
2p+2p+1p
4 coins:
1p+1p+1p+1p 2p+1p+1p+1p
1p+1p+1p+2p
1p+1p+2p+1p
1p+2p+1p+1p
5p:
1p+1p+1p+1p+1p
(9 of 25) [12/06/2001 17:13:40]

The mathematics of the Fibonacci series
If you count the number of solutions in each box, it will be exactly the form of Pascal's triangle that we showed above!

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