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Question-180746




Question Number 180746 by Acem last updated on 16/Nov/22
Answered by MJS_new last updated on 16/Nov/22
(1+(1/(200)))^(200) >2  because we know  (1+(1/1))^1 =2  (1+(1/n))^n =(((n+1)^n )/n^n ) is increasing  lim_(n→∞)  (1+(1/n))^n =e>2
$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{200}}\right)^{\mathrm{200}} >\mathrm{2} \\ $$$$\mathrm{because}\:\mathrm{we}\:\mathrm{know} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}}\right)^{\mathrm{1}} =\mathrm{2} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} =\frac{\left({n}+\mathrm{1}\right)^{{n}} }{{n}^{{n}} }\:\mathrm{is}\:\mathrm{increasing} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} =\mathrm{e}>\mathrm{2} \\ $$
Answered by mr W last updated on 17/Nov/22
(1+x)^n =1+nx+((n(n−1)x^2 )/2)+...>1+nx  with x=0.005 and n=200:  1.005^(200) >1+200×0.005=2 ✓  even:  1.005^(200) >1+200×0.005+(1/2)+(200×0.05)(200×0.05−0.005)                  =1+1+((1×0.995)/2)>2.44  even:   1.005^(200) >1+1+((1×0.995)/2)+((1×0.995×0.99)/6)>2.66
$$\left(\mathrm{1}+{x}\right)^{{n}} =\mathrm{1}+{nx}+\frac{{n}\left({n}−\mathrm{1}\right){x}^{\mathrm{2}} }{\mathrm{2}}+…>\mathrm{1}+{nx} \\ $$$${with}\:{x}=\mathrm{0}.\mathrm{005}\:{and}\:{n}=\mathrm{200}: \\ $$$$\mathrm{1}.\mathrm{005}^{\mathrm{200}} >\mathrm{1}+\mathrm{200}×\mathrm{0}.\mathrm{005}=\mathrm{2}\:\checkmark \\ $$$${even}: \\ $$$$\mathrm{1}.\mathrm{005}^{\mathrm{200}} >\mathrm{1}+\mathrm{200}×\mathrm{0}.\mathrm{005}+\frac{\mathrm{1}}{\mathrm{2}}+\left(\mathrm{200}×\mathrm{0}.\mathrm{05}\right)\left(\mathrm{200}×\mathrm{0}.\mathrm{05}−\mathrm{0}.\mathrm{005}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}+\mathrm{1}+\frac{\mathrm{1}×\mathrm{0}.\mathrm{995}}{\mathrm{2}}>\mathrm{2}.\mathrm{44} \\ $$$${even}: \\ $$$$\:\mathrm{1}.\mathrm{005}^{\mathrm{200}} >\mathrm{1}+\mathrm{1}+\frac{\mathrm{1}×\mathrm{0}.\mathrm{995}}{\mathrm{2}}+\frac{\mathrm{1}×\mathrm{0}.\mathrm{995}×\mathrm{0}.\mathrm{99}}{\mathrm{6}}>\mathrm{2}.\mathrm{66} \\ $$
Commented by manxsol last updated on 17/Nov/22
Commented by manxsol last updated on 17/Nov/22
the reasoning iz fine???
$${the}\:{reasoning}\:{iz}\:{fine}??? \\ $$

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