Question Number 226598 by fantastic2 last updated on 07/Dec/25
$$\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{5}} ={a}_{\mathrm{5}} {x}^{\mathrm{5}} +{a}_{\mathrm{4}} {x}^{\mathrm{4}} +{a}_{\mathrm{3}} {x}^{\mathrm{3}} ..+{a}_{\mathrm{1}} {x}+{a}_{\mathrm{0}} \\ $$$${find}\:{a}_{\mathrm{5}} +{a}_{\mathrm{3}} +{a}_{\mathrm{1}} {and}\:{a}_{\mathrm{4}} +{a}_{\mathrm{2}} +{a}_{\mathrm{0}} \\ $$
Answered by mr W last updated on 07/Dec/25
$${with}\:{x}=\mathrm{1}: \\ $$$$\mathrm{3}^{\mathrm{5}} =\left({a}_{\mathrm{5}} +{a}_{\mathrm{3}} +{a}_{\mathrm{1}} \right)+\left({a}_{\mathrm{4}} +{a}_{\mathrm{2}} +{a}_{\mathrm{0}} \right)={A}+{B}\:\:\:…\left({i}\right) \\ $$$${with}\:{x}=−\mathrm{1}: \\ $$$$\left(−\mathrm{1}\right)^{\mathrm{5}} =\left(−{a}_{\mathrm{5}} −{a}_{\mathrm{3}} −{a}_{\mathrm{1}} \right)+\left({a}_{\mathrm{4}} +{a}_{\mathrm{2}} +{a}_{\mathrm{0}} \right)=−{A}+{B}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\mathrm{3}^{\mathrm{5}} −\left(−\mathrm{1}\right)^{\mathrm{5}} =\mathrm{2}{A}\: \\ $$$$\Rightarrow{A}=\mathrm{122}\:\Rightarrow{B}={A}+\left(−\mathrm{1}\right)^{\mathrm{5}} =\mathrm{121} \\ $$$${i}.{e}.\:{a}_{\mathrm{5}} +{a}_{\mathrm{3}} +{a}_{\mathrm{1}} =\mathrm{122},\:\:{a}_{\mathrm{4}} +{a}_{\mathrm{2}} +{a}_{\mathrm{0}} =\mathrm{121} \\ $$
Commented by fantastic2 last updated on 07/Dec/25
$${great}! \\ $$