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Question Number 219408 by CrispyXYZ last updated on 24/Apr/25

a_1 =(1/2). a_n =(2/(n+3))+(1−(1/(n+3)))^2 a_(n−1) .  Find the maximum of a_n .

$${a}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}.\:{a}_{{n}} =\frac{\mathrm{2}}{{n}+\mathrm{3}}+\left(\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{3}}\right)^{\mathrm{2}} {a}_{{n}−\mathrm{1}} . \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{of}\:{a}_{{n}} . \\ $$

Answered by SdC355 last updated on 24/Apr/25

a_n =((c_1 +n(n+7)+12)/((n+3)^2 ))  a_1 =((c_1 +20)/(16))=(1/2)  c_1 =−12  a_n =((n(n+7))/((n+3)^2 )) , n=21 max a_n =((49)/(48))

$${a}_{{n}} =\frac{{c}_{\mathrm{1}} +{n}\left({n}+\mathrm{7}\right)+\mathrm{12}}{\left({n}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$${a}_{\mathrm{1}} =\frac{{c}_{\mathrm{1}} +\mathrm{20}}{\mathrm{16}}=\frac{\mathrm{1}}{\mathrm{2}}\:\:{c}_{\mathrm{1}} =−\mathrm{12} \\ $$$${a}_{{n}} =\frac{{n}\left({n}+\mathrm{7}\right)}{\left({n}+\mathrm{3}\right)^{\mathrm{2}} }\:,\:{n}=\mathrm{21}\:\mathrm{max}\:{a}_{{n}} =\frac{\mathrm{49}}{\mathrm{48}} \\ $$

Commented by universe last updated on 24/Apr/25

sir can you explain how to find you value of  a_n  ??

$${sir}\:{can}\:{you}\:{explain}\:{how}\:{to}\:{find}\:{you}\:{value}\:{of} \\ $$$${a}_{{n}} \:?? \\ $$

Answered by universe last updated on 24/Apr/25

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