Question Number 125585 by Dwaipayan Shikari last updated on 12/Dec/20
$$\frac{\left(\frac{\mathrm{1}}{\mathrm{1}!}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{2}!}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{3}!}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{4}!}\right)^{\mathrm{2}} +...\:}{\left(\frac{\mathrm{1}}{\mathrm{1}!}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}!}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{3}!}\right)^{\mathrm{2}} +......} \\ $$
Commented by Dwaipayan Shikari last updated on 12/Dec/20
$${Kindly}\:{tap}\:'+'\:{to}\:{post}\:{questions} \\ $$
Commented by khaki last updated on 12/Dec/20
$$\left(\mathrm{2x}^{\mathrm{2}} \:−\mathrm{4x}−\mathrm{6}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{plz}\:\mathrm{solve}\:\mathrm{this} \\ $$
Commented by Boucatchou last updated on 12/Dec/20
$${This}\:{is}\:{just}\:{a}\:{plynomial},\:{can}\:{you}\:{ask}\:{precise}\:{your}\:{question}? \\ $$
Commented by mr W last updated on 12/Dec/20
$${to}\:{khaki}\:{sir}: \\ $$$${please}\:{read}\:{the}\:{topmost}\:{post}\:\left({Q}.\mathrm{0}\right)\:{of} \\ $$$${the}\:{forum}\:{and}\:{study}\:{the}\:{help}\:{supplied} \\ $$$${there}!\:{you}\:{should}\:{not}\:{post}\:{your} \\ $$$${question}\:{in}\:{the}\:{thread}\:{of}\:{a}\:{question} \\ $$$${from}\:{other}\:{people}.\:{you}\:{should}\:{open} \\ $$$${a}\:{new}\:{post}\:{for}\:{your}\:{own}\:{question}! \\ $$
Commented by MathSh last updated on 14/Dec/20
$$=\frac{\mathrm{1}−\frac{\mathrm{1}}{{e}}}{{e}−\mathrm{1}}=\frac{\mathrm{1}}{{e}} \\ $$