Question Number 386 by 123456 last updated on 25/Jan/15
$${p}_{{n}+\mathrm{1}} ={p}_{{n}} +{q}_{{n}} \\ $$$${q}_{{n}+\mathrm{1}} ={p}_{{n}} {q}_{{n}} \\ $$$${p}_{\mathrm{0}} =\mathrm{2} \\ $$$${q}_{\mathrm{0}} =\mathrm{3} \\ $$$${f}\left({x}\right)=\frac{{p}_{\mathrm{0}} }{{q}_{\mathrm{0}} }+\frac{{p}_{\mathrm{1}} }{{q}_{\mathrm{1}} }{x}+\frac{{p}_{\mathrm{2}} }{{q}_{\mathrm{2}} }{x}^{\mathrm{2}} +\frac{{p}_{\mathrm{3}} }{{q}_{\mathrm{3}} }{x}^{\mathrm{3}} \\ $$$${f}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)=? \\ $$
Answered by prakash jain last updated on 27/Dec/14
$${p}_{\mathrm{1}} =\mathrm{5},{q}_{\mathrm{1}} =\mathrm{6} \\ $$$${p}_{\mathrm{2}} =\mathrm{11},{a}_{\mathrm{2}} =\mathrm{30} \\ $$$${p}_{\mathrm{3}} =\mathrm{41},{q}_{\mathrm{3}} =\mathrm{330} \\ $$$${f}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)=\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{5}}{\mathrm{6}}×\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{11}}{\mathrm{30}}×\frac{\mathrm{4}}{\mathrm{9}}+\frac{\mathrm{41}}{\mathrm{330}}×\frac{\mathrm{8}}{\mathrm{27}} \\ $$$$=\frac{\mathrm{110}×\mathrm{27}×\mathrm{2}+\mathrm{345}×\mathrm{10}+\mathrm{33}×\mathrm{44}+\mathrm{41}×\mathrm{8}}{\mathrm{27}×\mathrm{330}} \\ $$$$=\frac{\mathrm{5940}+\mathrm{3450}+\mathrm{1452}+\mathrm{328}}{\mathrm{27}×\mathrm{330}}=\frac{\mathrm{11170}}{\mathrm{8910}}=\frac{\mathrm{1117}}{\mathrm{891}} \\ $$