Misc 10 - Chapter 8 Class 11 Binomial Theorem

Transcript

Misc, 10 Find the expansion of (3x2 2ax + 3a2)3 using binomial theorem. We know that (a + b)n = nC0 an b0 + nC1 an 1 b1 + nC2 an 2 b2 + . . + nCn 1 a1 bn 1 + nCn a0 bn Hence (a + b)3 = 3C0 a3 b0 + 3C1 a2b1 + 3C2 a1 b2 + 3C3 a0 b3 = 3 ! 0! 3 0 ! a3 1 + 3! 1! 3 1 ! a2 b + 3! 2! 3 1 ! ab2 + 3! 3 ! 3 3 ! 1 b3 = 3! 3! a3 + 3 2! 2! a2 b + 3 2! 2! ab2 + 3! 3! a3 = a3 + 3a2b + 3b2a + b3 Now, (3x2 2ax + 3a2)3 = (3x2 (2ax 3a2))3 = (3x2 a(2x 3a))3 So, (3x2 2ax + 3a2)3 = (3x2 a(2x 3a))3 Putting a = 3x2 and b = a ( 2x 3a) in (1) (a + b)3 = a3 + 3a2b + 3b2a + b3 (3x a (2x 3a2))3 = (3x2)3 + 3 (3x2)2 ( a(2x 3a)) + 3 ( a(2x 3a))2 (3x2) + ( a(2x 3a))3 = 27x6 27x4 (a (2x 3a) + 9x2a2 (2x 3a)2 a3 (2x 3a)3 = 27x6 27x4 (2xa 3a2) + 9x2a2 ((2x)2 + (3a))2 2 (x) (3a)) a3 ((2x)3 (3a)3 3(2x)2 + (3a) + 3 (2x) (3a)2] = 27x6 54x59 + 81 x4a2 + 9x2a2 (4x2 + 9a2 12xa ) a3 (8x3 27a3 36x2a + 54xa2) = 27x6 54x5a + 81 x4a2 + 36x2a2 + 81x2a2 108x3a3 8x3 a3 + 27a6 + 36x2a4 54xa5 = 27x6 54ax5 + 81 a2x4 + 36a2x4 108x3a3 8x3 a3 54a5 x + 27a6 = 27x6 54ax5 + 117a2x4 116a3x3 54a5 x + 27a6 Thus, (3x2 2ax + 3a2)3 = 27x6 54ax5 + 117a2x4 116a3x3 54a5 x + 27a6