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Example 2 - Compute (98)^4 - Using Binomial Theorem - Teachoo

Example 2 - Chapter 8 Class 11 Binomial Theorem - Part 2
Example 2 - Chapter 8 Class 11 Binomial Theorem - Part 3

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Transcript

Example 2 Compute (98)5. (98)5 = (100 – 2)5 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)5 = = 5!/0!( 5 − 0)! a5 + 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 = 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!)/(2 ×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 (1000000) (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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.