Expert: Scotto - 9/10/2011

Question
Dear Prof Scotto

We have the following formulas

1.
(a + b)^2 = a^2 + 2ab + b^2

2.
(a − b)^2 = a^2 − 2ab + b^2

3.
(a + b)^3 = a^3 + b^3 + 3ab(a + b);

4.
(a − b)^3 = a^3 − b^3 − 3ab(a − b);

Question :

Up to what maximum value of n we can compute/deduce the formula ?

i.e

(a+b)^ n = ? where n= 1,2,3,4 .......
(a-b)^ n = ? where n= 1,2,3,4 .......

Examples :

can we compute/deduce the formula for

(a+b) ^ 10 = ?

(a-b) ^ 20 = ?

Thanks & Regards,
Prashant S Akerkar

Answer
The way to do it with that special triangle.
On the edges, the value is always 1.
In the middle, it is the sum of the element to the left and right in the row above.

For example, the 1st row is 1 1.  This says that x+1 to the 1st is 1x+1 = x+1.

The 2nd row is 1 2 1; note that 1 is on either edge and 2 = 1+1.  This says that
(x+1)^2 = 1x^2 + 2x + 1 { note the leading 1 is put there to emphasize it is off the table;
that will not be done in the future, but note it really is there).

The 3rd row is 1 3 3 1; again, 1 is on the edge and both 1+2 and 2+1 are the 3's in the middle.
Here, (x+1)^3 = x^3 + 3x^2 + 3x + 1.

The 4th row is 1 4 6 4 1; again, 1 is on either end, with 4=1+3, 6=3+3, and 4=3+1.
That means (x+1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1.

The 5th row is 1 5 10 10 5 1, since it has 1 at both end, 1+4=5, 4+6=10, 6+4=10, and 4+1=5.
Thus, (x+1)^5 = x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1/

It should also be noted that there is only one maximum in the middle for even rows
and two duplicate numbers in the middle for odd numbers.  For example, the 3rd row is 1 3 3 1,
the 5th row is 1, 5, 10, 10, 5, 1.  This is true for every odd row.

This is referred to as Pascal's triangle; the first several rows looks like
       1
      1 1
     1 2  1
    1 3  3  1
   1 4  6  4  1
  1 5 10 10  5  1
 1 6 15 20 15  6  1
1 7 21 35 35 32  7 1 { note besides the 1's, all numbers are divisible by 7 }
1 8 28 56 70 56 28 8 1

As has been said and can be seen, each element in the middle is the sum of the two above it.