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

$${a}_{\mathrm{1000}} =\frac{\mathrm{1}}{\mathrm{499501}}\:? \\ $$

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