Evaluate (1 01)5 + (0 99)5

1 015+0 995=1+ 015+1- 015 =C0515+C1514 01+C2513 012+C3512 013+C451 014+C55 01 +C0515-C1514 01+C2513 012-C3512 013+C451 014-C55 01 =2C0515+C2513 012+C451 014 =21+5!2!5-2! 0001+5!4!5-4! 00000001 =21+10 0001+5 00000001 =21+ 001+ 00000005 =2 0020001Hence,1 015+0 995=2 0020001

Evaluate (1.01)⁵ + (0.99)⁵

{(1.01)}^{5} + {(0.99)}^{5} = {(1 + . 01)}^{5} + {(1 - . 01)}^{5} = [{}^{5}C_{0} {(1)}^{5} + {}^{5}C_{1} {(1)}^{4} (. 01) + {}^{5}C_{2} {(1)}^{3} {(. 01)}^{2} + {}^{5}C_{3} {(1)}^{2} {(. 01)}^{3} + {}^{5}C_{4} (1) {(. 01)}^{4} + {}^{5}C_{5} (. 01)] + [{}^{5}C_{0} {(1)}^{5} - {}^{5}C_{1} {(1)}^{4} (. 01) + {}^{5}C_{2} {(1)}^{3} {(. 01)}^{2} - {}^{5}C_{3} {(1)}^{2} {(. 01)}^{3} + {}^{5}C_{4} (1) {(. 01)}^{4} - {}^{5}C_{5} (. 01)] = 2 [{}^{5}C_{0} {(1)}^{5} + {}^{5}C_{2} {(1)}^{3} {(. 01)}^{2} + {}^{5}C_{4} (1) {(. 01)}^{4}] = 2 [1 + \frac{5!}{2! (5 - 2)!} (. 0001) + \frac{5!}{4! (5 - 4)!} (. 00000001)] = 2 [1 + 10 (. 0001) + 5 (. 00000001)] = 2 (1 + . 001 + . 00000005) = 2.0020001 Hence, {(1.01)}^{5} + {(0.99)}^{5} = 2.0020001

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Evaluate (1.01)5 + (0.99)5

Evaluate (1.01)⁵ + (0.99)⁵