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Question Number 55408 by Tinkutara last updated on 23/Feb/19
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Feb/19
![a_(n+1) =a_n (a_n +1) (1/a_(n+1) )=(1/(a_n (a_n +1)))=(1/a_n )−(1/(a_n +1)) (1/(a_n +1))=(1/a_n )−(1/a_(n+1) ) (1/(a_2 +1))=(1/a_2 )−(1/a_3 ) (1/(a_3 +1))=(1/a_3 )−(1/a_4 ) (1/(a_4 +1))=(1/a_4 )−(1/a_5 ) ... ... (1/(a_n +1))=(1/a_n )−(1/a_(n+1) ) add S_★ =(1/a_2 )−(1/a_(n+1) ) a_1 =(1/3) a_2 =a_1 ^2 +a_1 =(1/9)+(1/3)=((1+3)/9)=(4/9) S_★ =(9/4)−(1/a_(n+1) ) S=(1/a_2 )+(1/a_3 )+(1/a_4 )+...+(1/a_n ) S_★ =(1/(a_2 +1))+(1/(a_3 +1))+..+(1/(a_n +1))=(9/4)−(1/a_(n+1) ) S>S_★ [S_★ ]=[(9/4)−(1/a_(n+1) )]=[2+(1/4)−(1/a_(n+1) )]=2 so S>S_★ S>2 it is not the answer of tbe question but somehow i tried to co−relate...](Q55437.png)
$${a}_{{n}+\mathrm{1}} ={a}_{{n}} \left({a}_{{n}} +\mathrm{1}\right) \\ $$$$\frac{\mathrm{1}}{{a}_{{n}+\mathrm{1}} }=\frac{\mathrm{1}}{{a}_{{n}} \left({a}_{{n}} +\mathrm{1}\right)}=\frac{\mathrm{1}}{{a}_{{n}} }−\frac{\mathrm{1}}{{a}_{{n}} +\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{{a}_{{n}} +\mathrm{1}}=\frac{\mathrm{1}}{{a}_{{n}} }−\frac{\mathrm{1}}{{a}_{{n}+\mathrm{1}} } \\ $$$$\frac{\mathrm{1}}{{a}_{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{{a}_{\mathrm{2}} }−\frac{\mathrm{1}}{{a}_{\mathrm{3}} } \\ $$$$\frac{\mathrm{1}}{{a}_{\mathrm{3}} +\mathrm{1}}=\frac{\mathrm{1}}{{a}_{\mathrm{3}} }−\frac{\mathrm{1}}{{a}_{\mathrm{4}} } \\ $$$$\frac{\mathrm{1}}{{a}_{\mathrm{4}} +\mathrm{1}}=\frac{\mathrm{1}}{{a}_{\mathrm{4}} }−\frac{\mathrm{1}}{{a}_{\mathrm{5}} } \\ $$$$... \\ $$$$... \\ $$$$\frac{\mathrm{1}}{{a}_{{n}} +\mathrm{1}}=\frac{\mathrm{1}}{{a}_{{n}} }−\frac{\mathrm{1}}{{a}_{{n}+\mathrm{1}} } \\ $$$${add} \\ $$$${S}_{\bigstar} =\frac{\mathrm{1}}{{a}_{\mathrm{2}} }−\frac{\mathrm{1}}{{a}_{{n}+\mathrm{1}} } \\ $$$${a}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${a}_{\mathrm{2}} ={a}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{1}+\mathrm{3}}{\mathrm{9}}=\frac{\mathrm{4}}{\mathrm{9}} \\ $$$${S}_{\bigstar} =\frac{\mathrm{9}}{\mathrm{4}}−\frac{\mathrm{1}}{{a}_{{n}+\mathrm{1}} } \\ $$$${S}=\frac{\mathrm{1}}{{a}_{\mathrm{2}} }+\frac{\mathrm{1}}{{a}_{\mathrm{3}} }+\frac{\mathrm{1}}{{a}_{\mathrm{4}} }+...+\frac{\mathrm{1}}{{a}_{{n}} } \\ $$$${S}_{\bigstar} =\frac{\mathrm{1}}{{a}_{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{{a}_{\mathrm{3}} +\mathrm{1}}+..+\frac{\mathrm{1}}{{a}_{{n}} +\mathrm{1}}=\frac{\mathrm{9}}{\mathrm{4}}−\frac{\mathrm{1}}{{a}_{{n}+\mathrm{1}} } \\ $$$${S}>{S}_{\bigstar} \\ $$$$\left[{S}_{\bigstar} \right]=\left[\frac{\mathrm{9}}{\mathrm{4}}−\frac{\mathrm{1}}{{a}_{{n}+\mathrm{1}} }\right]=\left[\mathrm{2}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{{a}_{{n}+\mathrm{1}} }\right]=\mathrm{2} \\ $$$${so}\:{S}>{S}_{\bigstar} \\ $$$${S}>\mathrm{2} \\ $$$${it}\:{is}\:{not}\:{the}\:{answer}\:{of}\:{tbe}\:{question}\:{but}\:{somehow} \\ $$$${i}\:{tried}\:{to}\:{co}−{relate}... \\ $$