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Question Number 19064 by 99 last updated on 03/Aug/17

Commented by 99 last updated on 03/Aug/17

sir please solve this question

$${sir}\:{please}\:{solve}\:{this}\:{question} \\ $$

Commented by 99 last updated on 03/Aug/17

question no.3

$${question}\:{no}.\mathrm{3} \\ $$

Answered by prakash jain last updated on 04/Aug/17

a_n =Σ_(i=1) ^(2^n −1) (1/i) 1+(1/2)+(1/3)+(1/4)+(1/5)+(1/6)+(1/7)+(1/8)+..+(1/(2^n −1)) >1+(1/2)+(1/4)+(1/4)+(1/8)+(1/8)+(1/8)+(1/8)+.. +(1/2^n )+(1/2^n )+..+(1/2^n )−(1/2^n ) =1+(1/2)+(1/2)+(1/2)+{n times}−(1/2^n ) =1+(n/2)−(1/2^n ) a(200)>100 −−−−−−−−−− 1+(1/2)+(1/3)+(1/4)+(1/5)+(1/6)+(1/7)+(1/8)+..+(1/(2^n −1)) <1+(1/2)+(1/2)+(1/4)+(1/4)+(1/4)+(1/4)+..+(1/2^(n−1) )+..+(1/2^(n−1) ) =1+n−1=n a(100)<100

$${a}_{{n}} =\underset{{i}=\mathrm{1}} {\overset{\mathrm{2}^{{n}} −\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{i}} \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{8}}+..+\frac{\mathrm{1}}{\mathrm{2}^{{n}} −\mathrm{1}} \\ $$$$>\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{8}}+.. \\ $$$$\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{2}^{{n}} }+\frac{\mathrm{1}}{\mathrm{2}^{{n}} }+..+\frac{\mathrm{1}}{\mathrm{2}^{{n}} }−\frac{\mathrm{1}}{\mathrm{2}^{{n}} } \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}+\left\{{n}\:{times}\right\}−\frac{\mathrm{1}}{\mathrm{2}^{{n}} } \\ $$$$=\mathrm{1}+\frac{{n}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}^{{n}} } \\ $$$${a}\left(\mathrm{200}\right)>\mathrm{100} \\ $$$$−−−−−−−−−− \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{8}}+..+\frac{\mathrm{1}}{\mathrm{2}^{{n}} −\mathrm{1}} \\ $$$$<\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}+..+\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }+..+\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$$$=\mathrm{1}+{n}−\mathrm{1}={n} \\ $$$${a}\left(\mathrm{100}\right)<\mathrm{100} \\ $$

Commented by rahul 19 last updated on 04/Oct/18

wow

$${wow} \\ $$

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