Page:Elementary algebra (1896).djvu/358

340 ALGEBRA. The last term in the product is bcd...k; denote it by S_n.

Hence (a+b)(a+c)(a+d)...(a+k) = a^n+ S_1 a^n-1+ S_2 a^n-2 +... + S_r a^n-r +...+S_n-1 a+ S_n.

In S_1, the number of terms is n; in S_2 the number of terms is the same as the number of combinations of n things two at a time; that is, nC2; in S_3 the number of terms is nC_3; and so on.

Now suppose c, d,... k, each equal to b; then S_1 becomes nC_1 b; S_2 becomes nC_2b^2; S_3 becomes nC_3b^3; and so on; thus

(a+b)^n= a^n + nC_1 b + nC_2 b^2 + nC_3 b^3 +... + nC_n b^n;

substituting for nC_1, nC_2,... we obtain

(a+b)^n =a^n + na^{n-1}b + n(n-1) 1 2 a^{n-2} b^2 + n(n-1)(n-2) 1 2 3 a^{n-3} b^3 +...+ b^n

the series containing n+1 terms.

This is the Binomial Theorem, and the expression on the right side is said to be the expansion of (a+b)^n.

409. The coefficients in the expansion of (a + n)^n are very conveniently expressed by the symbols {}^nC_1, {}^nC_2, {}^nC_3...{}^nC_n. We shall, however, sometimes further abbreviate them by omitting n, and writing C_1, C_2, C_3,...C_n. With this notation we have

(a+b)^n= a^n + C_1 b + C_2 b^2 + C_3 b^3 +... + C_n b^n;

If we write -b in the place of b, we obtain (a+b)^n= a^n + nC_1 (-b) + nC_2 (-b)^2 + nC_3 (-b)^3 +... + nC_n (-b)^n =a^n - C_1 b + C_2 b^2 - C_3 b^3 + ... + (-1)^n C_n b^n;

Thus the terms in the expansion of (a+b)^n and (a-b)^n are numerically the same, but in (a-b)^n they are alternately positive and negative, and the last term is positive or negative according as n is even or odd.