The formula for the kth term of (A + B)N is C(N,k-1)AN-k+1Bk-1 The eighth term of the expansion (2c+d)^7 So substitute A = 2c, B = d, k = 8, N = 7 C(N,k-1)AN-k+1Bk-1 C(7,8-1)(2c)7-8+1d8-1 C(7,7)(2c)0d7 (1)(1)d7 d7 The first term of the expansion of (3q-7r)^6 So substitute A = 3q, B = -7r, k = 1, N = 6 C(N,k-1)AN-k+1Bk-1 C(6,1-1)(3q)6-1+1(-7r)1-1 C(6,0)(3q)6(-7r)0 (1)(3q)6(1) (3q)6 36q6 729q6 Edwin