Binomial Theorem

 Statement: When n is a positive integer,

             (a + x)n = an + C(n, 1)an -1x + C(n, 2)an -2x2 +…. + C(n, r)an - r xr +…+ C(n, n)xn

 Proof:  By actual multiplication, we have

             (a + x)2 = a2 + 2ax + x2 = a2 + C(n, 1) ax + C(2, 2) x2

             (a + x)3 = a3 + 3a2x + 3ax2 + x3

                          = a3 + C(3, 1) a2x + C(3, 2) ax2 + C(3, 3) x3

          The theorem is thus seen to be true when n = 2 and 3.

          Let us assume therefore that the theorem is true when n has some particular value, say m, i.e.(a + x)m = am + C(m, 1)am-1x + C(m, 2)am-2 x2+ …+ C(m, r)am - rxr + …+ 

                    C(m, m)xn                              ….(i)

        Multiplying both sides of (i) by (a + x), we have

(a+x)m + 1= (a + x){am + C(m, 1)am-1x+ C(m, 2)am-2 x2 +…+ C(m, r)am - r xr +…+ C(m, m)xn}

= am + 1 + [C(m, 1) + C(m, 0)] amx + [C(m, 2) + C(m, 1)] am - 1 x2 + [C(m, r) + C(m, r - 1)] + … + xm + 1

             Since C(m, r) + C(m, r - 1) = C(m + 1, r) and  C(m, 1) + C(m, 0)

                                                          = C(m + 1, 1)

             \   (a +x)m + 1= am + 1 + C(m + 1, 1)am x + C(m+1, 2)am - 1 x2 +…

             + C(m + 1, r)am - r + 1xr +… + xm + 1

             Thus the theorem is true for n = m, it is also true for n = m + 1. But it is true for n = 3, therefore it is true for n = 4 and being true for n = 4 it must be true for n = 5 and so on. Hence, the theorem is true for all positive integral value of n.

Note:

1.          When the quantity is expressed in a series, it is said to be expanded and the right side of above relation is called the expansion of (a + x)n and the coefficients C(n, 0), C(n, 1), C(n, 2), … C(n, r), … C(n, n) of the successive terms are called binomial coefficients.

2.          It is clear from the above expansion that the number of terms in the expansion of (a + x)n is finite and is equal to (n + 1), i.e., one more than the index of (a + x).

3.          In every term the index of x is always one less than ordinal number of the term and the sum of the indices of a and x is n.

4.         If we put a = 1, then the binomial expansion will be

            (1 + x)n = 1 + C(n, 1) x + C(n, 2)x2 + C(n, 3)x3 + … + C(n, r)xr + … + xn.

General Term

            The (r + 1)th term in the expansion of (a + x)n is usually called its general term, because any required term may be found from it, by giving a suitable value of r. The (r + 1)th term is generally denoted by tr + 1.

            In the expansion of (a + x)n

             1st term t1 = C(n, 0) an

             2nd term t2 = C(n, 1) an -1 x

             3rd term t3 = C(n, 2) an - 2x2

             …        ….        …

(r + 1)th term tr + 1  = C(n, r) an - r xr

Similarly, in the expansion of (1 + x)n the (r + 1)th term = C(n, r) xr