# Example 2 - Chapter 8 Class 11 Binomial Theorem

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 2 Compute (98)5. (98)5 = (100 – 2)5 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)5 = = 5!0! 5 − 0! a5 × 1 + 5!1! 5 − 1! a4 b1 + 5!2! 5 − 2! a3 b2 + 5!3! 5 − 3! a2b3 + 5!4! 5 − 4! a b4 + 5!5! 5 −5! b5 × 1 = 5!0! × 5! a5 + 5!1! × 4! a4 b + 5!2! 3! a3 b2 + 5!3! 2! a2b3 + 5!4! 1! a b4 + 5!5! 0! b5 = 5!5! a5 + 5 × 4!4! a4 b + 5 × 4 × 3!2! 3! a3 b2 + 5 × 4 × 3!2 × 1 ×3! a3b2 + 5 × 4 × 3!3! ×1 ×3! a2b3 + 5 × 4!4! ab4 + 5!5! b5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 Thus, (a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 We need to find (100 – 2)5 Putting a = 100 & b = –2 (100 – 2)5 = (100)5 + 5(100)4 (–2) + 10(100)3 (–2)2 + 10 (100)2 (–2)3 + 5 (100) (–2)4 + (–2)5 (98)5 = 10000000000 + 5 (100000000) (–2) + 10 (10002000) (4) + 10(10000) (–8) + 5(100) (16) + (–32) = 10000000000 – 1000000000 + 40000000 – 800000 + 8000 – 32 = 9039207968 So, (98)5 = 9039207968

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.