1 5 10 10 5 1 Each coefficient of any row is obtained by adding two coefficients in the preceding row, one on the immediate left and the other on the immediate right and each row is bounded by 1 on both sides. The (r + 1)th term or general term is given by Tr + 1 = n Cr an – r br 8.1.4 Some particular cases If n is a positive integer, then (a + b)n = nC0 an b0 + nC1 an b1 + nC2 an – 2 b2 + ... + nCr an – r br + ... + Cn a0 bn ... (1) In particular 1. Replacing b by – b in (i), we get (a – b)n = nC0 an b0 – nC1 an – 1 b1 + nC2 an – 2 b2 + ... + (–1)r nCr an – r br + ... + (–1)n nCn a0 bn ... (2) 2. Adding (1) and (2), we get (a + b)n + (a – b)n = 2 [nC0 an b0 + nC2 an – 2 b2 + nC4 an – 4 b4 + ... ] = 2 [terms at odd places] 3. Subtracting (2) from (1), we get (a + b)n – (a – b)n = 2 [nC1 an – 1 b1 + nC3 an – 3 b3 + ... ] = 2 [sum of terms at even places] 4. Replacing a by 1 and b by x in (1), we get (1 + x)n =nC0 x0 + nC1 x + nC2 x2 + ... + nCr xr + ... + nCn – 1 xn – 1 + nCn xn i.e. (1 + x)n = Cr x r r =0 BINOMIAL THEOREM 131 5. Replacing a by 1 and b by –x in ... (1), we get (1 – x)n = nC0 x0 – nC1 x + nC2 x2 ... + nCn–1 (–1)n–1 xn-1 + nCn (–1)n xn i.e., (1 – x)n = (−1) r n Cr x r r =0 8.1.5 The pth term from the end The pth term from the end in the expansion of (a + b) n is (n – p + 2)th term from the beginning. 8.1.6 Middle terms The middle term depends upon the value of n.
- (a)If n is even: then the total number of terms in the expansion of (a + b)n is n + 1 (odd). Hence, there is only one middle term, i.e., n term is the middle +1 term.
- (b)If n is odd: then the total number of terms in the expansion of (a + b)n is n + 1 th th n +1 n+3 (even). So there are two middle terms i.e., and are two 2 2 middle terms. 8.1.7 Binomial coefficient In the Binomial expression, we have (a + b)n =nC0 an + nC1 an – 1 b + nC2 an – 2 b2 + ... + nCn bn ... (1) n n n n The coefficients C0, C1, C2, ... , Cn are known as binomial or combinatorial coefficients. Putting a = b = 1 in (1), we get C0 + nC1 + nC2 + ... + nCn = 2n Thus the sum of all the binomial coefficients is equal to 2n. Again, putting a = 1 and b = –1 in (i), we get C0 + nC2 + nC4 + ... = nC1 + nC3 + nC5 + ... Thus, the sum of all the odd binomial coefficients is equal to the sum of all the even 2n binomial coefficients and each is equal to = 2 n−1 . C0 + nC2 + nC4 + ... = nC1 + nC3 + nC5 + ... = 2n – 1