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Our first topic is the Pascal Triangle, an infinite array of natural numbers. We begin by considering expansions of the powers (a + b)n of a sum of two terms. Clearly, (a + b)2 = (a + b)(a + b) = a2 + ab + ba + b2 = a2 + 2ab + b2. Then (a + b)3 = (a+b)2(a+b) = (a2 + 2ab + b2)(a + b). We expand this last expression as the sum of all products of a term of a2 + 2ab + b2 by a term of a + b in the following manner:
a2 + 2ab + b2 a + b a3 + 2a2b + ab2 a2b + 2ab2 + b3 a3 + 3a2b + 3ab2 + b3Hence (a + b)3= a3 + 3a2b + 3ab2 + b3. If
and 
(a is not equal to zero and b is not equal to zero), this
may be written (a + b)3 = a3b0
+ 3a2b1 + 3a1b2
+ a0b3.
The terms of the expanded form are such that the exponent for a starts as 3 and decreases by one each time, while the exponent of b starts as 0 and increases by one each time. Thus the sum of the exponents is 3 in each term.
One might guess that by analogy the expansion of (a + b)4 involves a4, a3b, a2b2, ab3, and b4. This is verified by expanding (a+b)4 = (a + b)3(a + b) = (a3 + 3a2b + 3ab2 + b3)(a + b) as follows:
(1) a3 + 3a2b + 3ab2 + b3 (2) a + b (3) a4 + 3a3b + 3a2b2 + ab3 (4) a3b + 3a2b2 + 3ab3 + b4 (5) a4 + 4a3b + 6a2b2 + 4ab3 + b4Thus we see that a4, a3b, a2b2, ab3, and b4 are multiplied by 1, 4, 6, 4, 1 to form the terms of the expansion. The numbers 1, 4, 6, 4, 1 are the coefficients of the expansion. Examination of expressions (1) to (5), above, shows that these coefficients are obtainable from the coefficients 1, 3, 3, 1 of (a + b)3 by means of the following condensed versions of (3), (4), and (5):
(3*) 1 3 3 1 (4*) 1 3 3 1 (5*) 1 4 6 4 1
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