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प्रश्न
Use binomial theorem to evaluate the following upto four places of decimal
(0.98)–3
उत्तर
(0.98)–3
= (1 – 0.02)–3
= `1 - (-3) (0.02) + ((-3)(-3 - 1))/(2!) (0.02)^2 - ((-3)(-3 - 1)(-3 - 2))/(3!) (0.02)^3 + ......`
= `1 + 0.06 + (-3(-4))/2 (0.02)^2 - ((-3)(-4)(-5))/6 (0.02)^3 + ......`
= 1 + 0.06 + 0.0024 + 0.00008 + ….
= 1.06248
= 1.0625
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