Ex 8.1,3 - Chapter 8 Class 11 Binomial Theorem

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Ex8.1, 3 Expand the expression (2x 3)6 (2x 3)6 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)6 = = 6!/(0! (6 0)!) a6 1 + 6!/(1 ! (6 1) !) a5 b + 6!/2!(6 2)! a4 b2 + 6!/(3 !(6 3)!) a3 b3 + 6!/(4 ! (6 4 ) !)a2 b4 + 6!/(5 !( 6 5)!) ab5 + 6!/(6 ! (6 6) !) 1 b6 = 6!/(1 6! ) a6 + 6!/(1 5!) a5 b + 6!/(2! 4!) a4 b2 + 6!/(3 ! 3!) a3 b3 + 6!/(4 ! 2!) a2 b4 + 6!/(5 ! 1) a b5 + 6!/(6 ! 1) b6 = 6!/6! a6 + (6 5!)/(5! ) a5b + (6 5 4!)/(2 4!) a4 b2 + (6 5 4 3!)/(3 2 1 3!) a3 b3 + (6 5 4!)/(2 1 4!) a2 b4 + (6 5!)/(1 5!) ab5 + 6!/6! b6 = a6 + 6a5b + (3 5)a4b2 + (5 4)a3b3 + (3 5)a2b4 + 6ab5 + b6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6 Thus, (a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6 We need to find (2x 3)6 Putting a = 2x & b = 3 (2x 3)6 = (2x)6 + 6 (2x)5 ( 3)2 + 15 (2x)4 ( 3) + 20 (2x)3 ( 3)3 + 15 (2x)2 ( 3)+ 6 (2x) ( 3)5 + ( 3)6 = 64x6 18(32x5) + 15(16x4) (9) + 20 (8x3) ( 27) + 15 (4x2) (18) + 6 (2x) ( 243) + (729) = 64x6 576x5 + 2160x4 4320x3 + 4860x2 2916x + 729