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Question-82995




Question Number 82995 by M±th+et£s last updated on 26/Feb/20
Commented by mr W last updated on 26/Feb/20
a_(1000) =(1/(499501)) ?
$${a}_{\mathrm{1000}} =\frac{\mathrm{1}}{\mathrm{499501}}\:? \\ $$
Answered by mind is power last updated on 26/Feb/20
⇒(1/a_(n+1) )=n+(1/a_n )  let b_n =(1/a_n )  ⇒b_(n+1) =n+b_n   b_0 =1  ⇒Σ_(k=0) ^(n−1) (b_(k+1) −b_k )=Σ_(k=0) ^(n−1) k  ⇒b_n −b_0 =(n/2)(n−1)  ⇒b_n =1+(n/2)(n−1)  ⇒a_n =(1/b_n )=(2/(n^2 −n+2))  a_(1000) =(2/(999002))=(1/(499501))
$$\Rightarrow\frac{\mathrm{1}}{{a}_{{n}+\mathrm{1}} }={n}+\frac{\mathrm{1}}{{a}_{{n}} } \\ $$$${let}\:{b}_{{n}} =\frac{\mathrm{1}}{{a}_{{n}} } \\ $$$$\Rightarrow{b}_{{n}+\mathrm{1}} ={n}+{b}_{{n}} \\ $$$${b}_{\mathrm{0}} =\mathrm{1} \\ $$$$\Rightarrow\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left({b}_{{k}+\mathrm{1}} −{b}_{{k}} \right)=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{k} \\ $$$$\Rightarrow{b}_{{n}} −{b}_{\mathrm{0}} =\frac{{n}}{\mathrm{2}}\left({n}−\mathrm{1}\right) \\ $$$$\Rightarrow{b}_{{n}} =\mathrm{1}+\frac{{n}}{\mathrm{2}}\left({n}−\mathrm{1}\right) \\ $$$$\Rightarrow{a}_{{n}} =\frac{\mathrm{1}}{{b}_{{n}} }=\frac{\mathrm{2}}{{n}^{\mathrm{2}} −{n}+\mathrm{2}} \\ $$$${a}_{\mathrm{1000}} =\frac{\mathrm{2}}{\mathrm{999002}}=\frac{\mathrm{1}}{\mathrm{499501}} \\ $$$$ \\ $$
Commented by mr W last updated on 26/Feb/20
fine sir! i got the same.
$${fine}\:{sir}!\:{i}\:{got}\:{the}\:{same}. \\ $$
Commented by M±th+et£s last updated on 26/Feb/20
thank you sir.this is the value
$${thank}\:{you}\:{sir}.{this}\:{is}\:{the}\:{value} \\ $$
Commented by mind is power last updated on 27/Feb/20
withe pleasur
$${withe}\:{pleasur} \\ $$

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