Ex 8.1,9 - Chapter 8 Class 11 Binomial Theorem

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Ex 8.1,9 Using Binomial Theorem, evaluate (99)5 (99)5 = (100 1)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 1)5 Putting a = 100 & b = 1 (100 1)5 = (100)5 + 5(100)4 ( 1) + 10 (100)3 ( 1)2 + 10(100)2 ( 1)3 + 5(100) ( 1)4 + ( 1)5 (99)5 = 10000000000 5(100000000) + 10(1000000) 10(10000) + 5 (100) 1 = 10000000000 500000000 + 100000 100000 + 500 1 = 9509900499