Ex 8.1,3 - Chapter 8 Class 11 Binomial Theorem (Deleted)
Last updated at Jan. 29, 2020 by Teachoo
Last updated at Jan. 29, 2020 by Teachoo
Transcript
Ex 8.1, 3 Expand the expression (2x – 3)6 (2x – 3)6 We know that (a + b)n = nC0 an + nC1 an – 1 b1 + nC2 an – 2 b2 + ….…. + nCn – 1 a1 bn – 1 + nCn bn + nCn a0 bn Hence (a + b)6 = = 6!/(0! (6 −0)!) a6 + 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) !) 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 + 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) + 15 (2x)4 (–3)2 + 20 (2x)3 (–3)3 + 15 (2x)2 (–3)4 + 6 (2x) (–3)5 + (–3)6 = 64x6 – 18(32x5) + 15(16x4) (9) + 20 (8x3) (–27) + 15 (4x2) (81) + 6 (2x) (–243) + (729) = 64x6 – 576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729
Ex 8.1
Ex 8.1,2 Important Deleted for CBSE Board 2022 Exams
Ex 8.1,3 Deleted for CBSE Board 2022 Exams You are here
Ex 8.1,4 Important Deleted for CBSE Board 2022 Exams
Ex 8.1, 5 Deleted for CBSE Board 2022 Exams
Ex 8.1 6 Deleted for CBSE Board 2022 Exams
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Ex 8.1,10 Important Deleted for CBSE Board 2022 Exams
Ex 8.1,11 Deleted for CBSE Board 2022 Exams
Ex 8.1,12 Important Deleted for CBSE Board 2022 Exams
Ex 8.1,13 Important Deleted for CBSE Board 2022 Exams
Ex 8.1,14 Important Deleted for CBSE Board 2022 Exams
Ex 8.1
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