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प्रश्न
Evaluate the following using binomial theorem:
(999)5
उत्तर
(x + a)n = nC0 xn a0 + nC1 xn-1 a1 + nC2 xn-2 a2 + ……… + nCr xn-r ar + …… + nCn an
(999)5 = (1000 – 1)5 = 5C0 (1000)5 – 5C1 (1000)4 (1)1 + 5C2 (1000)3 (1)2 – 5C3 (1000)2 (1)3 + 5C4 (1000)5 (1)4 – 5C5 (1)5
= 1(1000)5 – 5(1000)4 – 10(1000)3 – 10(1000)2 + 5(1000) – 1
= 1000000000000000 – 5000000000000 + 10000000000 – 10000000 + 5000 – 1
= 995009990004999
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