Find the value of $(1.01)^{10} + (1 - 0.01)^{10}$ correct to 7 places of decimal.

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Solution

$(1.01)^{10} + (1 - 0.01)^{10} = (1 + 0.01)^{10} + (1 - 0.01)^{10}$

$= (^{10} C_{1} +^{10} C_{2} \frac{1}{10^{2}} +^{10} C_{3} \frac{1}{10^{3}} . . . . . +^{10} C_{10} \frac{1}{10^{10}}) + (^{10} C_{1} -^{10} C_{2} \frac{1}{10^{2}} +^{10} C_{3} \frac{1}{10^{3}} -^{10} C_{3} \frac{1}{10^{3}} -^{10} C_{4} \frac{1}{10^{4}})$

$= 2 (^{10} C_{1} -^{10} C_{3} \frac{1}{10^{3}} +^{10} C_{5} \frac{1}{10^{5}} +^{10} C_{7} \frac{1}{10^{7}} +^{10} C_{9} \frac{1}{10^{9}})$

$= 2 (10 + \frac{10!}{3! 7!} \frac{1}{1000} + \frac{10!}{5! 5} \frac{1}{(10)^{5}} + \frac{10!}{7! 3!} \times \frac{1}{10^{7}} + \frac{10!}{9! 1!} \frac{1}{10^{9}})$

$= 2 (10 + \frac{9 \times 8}{3 \times 2 \times 1000} + \frac{9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 10^{5}} + \frac{9 \times 8}{3 \times 2 \times 10^{7}} + \frac{1}{10^{8}})$

=2.0090042

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