BINOMIAL THEOREM. 343
PROOF BY MATHEMATICAL INDUCTION.
413. By actual multiplication we obtain the following identities : (a+ b)2 = a2 + 2ab +b2, (a+ b)3 = a3+ 3a2b + 3ab2 + b3, (a+ b)4 =a + 4a3b + 6a2b2 + 4ab3 + b4.
Selecting any one of these, and rewriting so as to exhibit the laws of formation of exponents and coefficients, we have
(a+b)4 =a4 + 4 1 a4-1b + 4 3 1 2 a4-2b2 + 4 3 2 1 2 3 a4-3b3 + 4 3 2 1 1 2 3 4a0b4 (Art. 216).
If these laws of formation hold for (a + b)n, n being any positive integer, then (a+b)n = an + nan-1b + n(n-1) 1 2 an-2 b2 + n(n-1)(n-2) 1 2 3 an-3 b3 + ... (1)
Multiplying each side of the assumed identity by (a+b) and combining terms, we obtain
(a+b)n+1 = an+1 + nan-1b + n(n+1) 1 2 an-1 b2 + n(n+1)(n-1) 1 2 3 an-2 b3 + (2)
It will be seen that n in (1) is, in every instance, replaced by (n+1) in (2). Hence if the theorem be true for any value of n, it will be true for the next higher value. We have shown by multiplication that the theorem is true when nm successively equals 2, 3, and 4; hence it is true when n=5, and so on indefinitely. The theorem is therefore true for all positive integral values of n.
