Ex 8.1,12 - Chapter 8 Class 11 Binomial Theorem (Deleted)
Last updated at Sept. 3, 2021 by Teachoo
Ex 8.1
Ex 8.1,2 Important Deleted for CBSE Board 2022 Exams
Ex 8.1,3 Deleted for CBSE Board 2022 Exams
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
Ex 8.1,7 Important Deleted for CBSE Board 2022 Exams
Ex 8.1,8 Deleted for CBSE Board 2022 Exams
Ex 8.1,9 Deleted for CBSE Board 2022 Exams
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 You are here
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
Last updated at Sept. 3, 2021 by Teachoo
Ex 8.1, 12 Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate (√2+1)^6 + (√2−1)^6. We know that (a + b)n = nC0 an + nC1 an – 1 b1 + nC2 an – 2 b2 + ….…. + nCn – 1 a1 bn – 1 + nCn 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) !) 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 For (a – b)6 Replace b by (–b) in (1) (a + (–b)) 6 = a6 + 6a5 (–b) + 15a4 (– b)2 + 20a3 (– b)3 + 15a2 (– b)4 + 6a(– b)5 + (– b)6 (a – b)6 = a6 – 6a5 b + 15a4 b2 – 20a3 b3 + 15a2 b4 – 6ab5 + b6 Now, (a + b)6 + (a – b)6 = (a6 + 6a5 b + 15a4 b2 + 20a3 b3 + 15a2 b4 + 6ab5 + b6) + (a6 – 6a5 b + 15a4 b2 – 20a3 b3 + 15a2 b4 – 6ab5 + b6) = a6 + a6 + 15a4 b2 + 15a4 b2 + 15a2 b4 + 15a2 b4 + b6 + b6 + 6a5 b – 6a5b + 20a3 b3 – 20a3 b3 + 6ab5 – 6ab5 = 2a6 + 30a6 b2+ 30a2 b4 + 2b6 + 0 + 0 + 0 = 2a6 + 30a6 b2+ 30a2 b4 + 2b6 = 2(a6 + 15a4 b2 + 15a2 b4 + b6 ) Thus, (a + b)6 + (a – b)6 = 2(a6 + 15a4 b2 + 15a2 b4 + b6 ) Putting a = x and b = 1 (x + 1)6 + (x – 1)6 = 2(x6 + 15x4 (1)2 + 15x2 (1)4 + (1)6 ) = 2(x6 + 15x4 + 15x2 + 1) = 2 ("(" 2")3 + 15(" 2")2 + 15 (" 2") + 1" ) = 2 (8 + 60 + 30 + 1) = 2 ( 99) = 198 Thus, (√2 + 1)6 + (√2– 1 )6 = 198 Ex 8.1,12 Find (x + 1)6 + (x 1)6. Hence or otherwise evaluate 2 +1 6 + 2 1 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 For (a b)6 Replace b by ( b) in (1) (a + ( b)) 6 = a6 + 6a5 ( b) + 15a4 ( b)2 + 20a3 ( b)3 + 15a2 ( b)4 + 6a( b)5 + ( b)6 (a b)6 = a6 6a5 b + 15a4 b2 20a3 b3 + 15a2 b4 6ab5 + b6 Now, (a + b)6 + (a b)6 = (a6 + 6a5 b + 15a4 b2 + 20a3 b3 + 15a2 b4 + 6ab5 + b6) + (a6 6a5 b + 15a4 b2 20a3 b3 + 15a2 b4 6ab5 + b6) = a6 + a6 + 15a4 b2 + 15a4 b2 + 15a2 b4 + 15a2 b4 + b6 + b6 + 6a5 b 6a5b + 20a3 b3 20a3 b3 + 6ab5 6ab5 = 2a6 + 30a6 b2+ 30a2 b4 + 6b6 + 0 + 0 + 0 = 2a6 + 30a6 b2+ 30a2 b4 + 6b6 = 2(a6 + 15a4 b2 + 15a2 b4 + b6 ) Thus, (a + b)6 + (a b)6 = 2(a6 + 15a4 b2 + 15a2 b4 + b6 ) Putting a = x and b = 1 (x + 1)6 + (x 1)6 = 2(x6 + 15x4 (1)2 + 15x2 (1)4 + (1)6 ) = 2(x6 + 15x4 + 15x2 + 1) Thus, (x + 1)6 + (x 1)6 = 2(x6 + 15x4 + 15x2 + 1) Putting x = 2 ( 2 + 1)6 + ( 2 1 )6 = 2 2 6 + 15 2 4 + 15 2 2 + 1 = 2 2 1 2 6 + 15 2 1 2 4 + 15 2 1 2 2 + 1 = 2 (2)3 + 15(2)2 + 15 (2) + 1 = 2 (8 + 60 + 30 + 1) = 2 ( 99) = 198 Thus, ( 2 + 1)6 + ( 2 1 )6 = 198