Question
Using binomial theorem determine which number is smaller $(1.2)^{4000}$ or $800?$
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Answer
$(1.2)^{4000}=(1+0.2)^{4000}$
$={^{4000}\text{C}}_0(0.2)^0(1)^{4000}+{^{4000}\text{C}}_1(0.2)^1{1}^{3999}+....\\+{^{4000}\text{C}}_400(0.2)^0(1)^{4000}(0.2)^{4000}1^0$
$=1 + 4000 \times 0.2 \times 1 +.......+(0.2)^{4000}$
$= 1 + 800 +.......+(0.2)^{4000}$
Here, we dearty observe $(1, 2)^{4000} $ is less than $(801)$ thus, $(1, 2)^{4000} < 800.$
$={^{4000}\text{C}}_0(0.2)^0(1)^{4000}+{^{4000}\text{C}}_1(0.2)^1{1}^{3999}+....\\+{^{4000}\text{C}}_400(0.2)^0(1)^{4000}(0.2)^{4000}1^0$
$=1 + 4000 \times 0.2 \times 1 +.......+(0.2)^{4000}$
$= 1 + 800 +.......+(0.2)^{4000}$
Here, we dearty observe $(1, 2)^{4000} $ is less than $(801)$ thus, $(1, 2)^{4000} < 800.$
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