Question Number 195898 by Calculusboy last updated on 12/Aug/23
Answered by qaz last updated on 13/Aug/23
$${a}_{{n}+\mathrm{2}} {a}_{{n}+\mathrm{1}} −{a}_{{n}+\mathrm{1}} {a}_{{n}} =\mathrm{2}\:\:\:\:\Rightarrow{a}_{{n}+\mathrm{2}} {a}_{{n}+\mathrm{1}} =\mathrm{2}{n}+{a}_{\mathrm{1}} {a}_{\mathrm{2}} =\mathrm{2}\left({n}+\mathrm{1}\right) \\ $$$$\frac{{a}_{{n}} +\frac{\mathrm{1}}{{a}_{{n}+\mathrm{1}} }}{{a}_{{n}+\mathrm{2}} }=\mathrm{1}−\frac{\mathrm{1}}{{a}_{{n}+\mathrm{2}} {a}_{{n}+\mathrm{1}} } \\ $$$$\mathrm{2}^{\alpha} \begin{pmatrix}{\mathrm{61}}\\{\mathrm{31}}\end{pmatrix}=\underset{{i}=\mathrm{1}} {\overset{\mathrm{30}} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{{a}_{{i}+\mathrm{2}} {a}_{{i}+\mathrm{1}} }\right)=\underset{{i}=\mathrm{1}} {\overset{\mathrm{30}} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\right) \\ $$$$=\underset{{i}=\mathrm{2}} {\overset{\mathrm{31}} {\prod}}\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{n}}=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{60}} }\begin{pmatrix}{\mathrm{61}}\\{\mathrm{31}}\end{pmatrix}\:\:\:\Rightarrow\alpha=−\mathrm{60} \\ $$
Commented by Calculusboy last updated on 13/Aug/23
$${thanks} \\ $$