Solution :
We have, \((1.01)^{1000000}\) – 10000
= \((1 + 0.01)^{1000000}\) – 10000
By using binomial theorem,
= \(^{1000000}C_0\) + \(^{1000000}C_1 (0.01)\) + \(^{1000000}C_2 (0.01)^2\) + …… + \(^{1000000}C_{1000000} (0.01)^{1000000}\) – 10000
= (1 + 1000000(0.01) + other positive terms) – 10000
= (1 + 10000 + other positive terms) – 10000
= 1 + other positive terms > 0
Hence, \((1.01)^{1000000}\) is greater than 10000.
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