Using Binomial Theorem, evaluate (102)5
Question:

Using Binomial Theorem, evaluate $(102)^{5}$

Solution:

102 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.

It can be written that, 102 = 100 + 2

$\therefore(102)^{5}=(100+2)^{5}$

$={ }^{5} \mathrm{C}_{0}(100)^{5}+{ }^{5} \mathrm{C}_{1}(100)^{4}(2)+{ }^{5} \mathrm{C}_{2}(100)^{3}(2)^{2}+{ }^{5} \mathrm{C}_{3}(100)^{2}(2)^{3}$

$+{ }^{5} \mathrm{C}_{4}(100)(2)^{4}+{ }^{5} \mathrm{C}_{5}(2)^{5}$

$=(100)^{5}+5(100)^{4}(2)+10(100)^{3}(2)^{2}+10(100)^{2}(2)^{3}+5(100)(2)^{4}+(2)^{5}$

$=10000000000+1000000000+40000000+800000+8000+32$

$=11040808032$

 

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