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Question Number 198813 by mnjuly1970 last updated on 24/Oct/23
suppose that : E= (√( 9 + 4 ( Σ_(k=2) ^n a_( k) ^( 2) ))) is given let a_( k) = a_(k−1) . (a_( k−1) +1) and a_1 =1 , a_( 2) = 2 , a_( 3) = 6 , a_4 = 42 , a_( 5) = 1806 and etc find the value of E =?
$$ \\ $$$$\:\:\:\:\:{suppose}\:\:{that}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\:{E}=\:\sqrt{\:\mathrm{9}\:+\:\mathrm{4}\:\left(\:\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\:\:{a}_{\:{k}} ^{\:\mathrm{2}} \:\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{is}\:\:{given} \\ $$$$\:\:\:\:\:{let}\:\:\:\:{a}_{\:{k}} \:=\:{a}_{{k}−\mathrm{1}} \:.\:\left({a}_{\:{k}−\mathrm{1}} \:+\mathrm{1}\right) \\ $$$$\:\:\:\:\:{and}\:\:\:{a}_{\mathrm{1}} \:=\mathrm{1}\:,\:\:\:{a}_{\:\mathrm{2}} \:=\:\mathrm{2}\:\:,\:{a}_{\:\mathrm{3}} =\:\mathrm{6}\:, \\ $$$$\:\:\:\:\:\:{a}_{\mathrm{4}} \:=\:\:\mathrm{42}\:\:\:,\:{a}_{\:\mathrm{5}} \:=\:\mathrm{1806}\:\:\:\:{and}\:\:{etc} \\ $$$$\:\:\:{find}\:{the}\:\:{value}\:\:{of}\:\:\:\:{E}\:=? \\ $$
Answered by witcher3 last updated on 24/Oct/23
a_k −a_(k−1) =a_(k−1) ^2 ⇔Σ_(k=3) ^(n+1) (a_k −a_(k−1) )=Σ_(k=2) ^n a_k ^2 ⇔a_(n+1) −a_2 =Σ_(k=2) ^n (a_k )^2 E=(√(9+4(a_(n+1) −a_2 )))=(√(9+4(a_(n+1) −2))) =(√(1+4a_(n+1) ))=(√(1+4a_n (a_n +1)))=(√((2a_n +1)^2 )) =∣2a_n +1∣=2a_n +1,a_n >0
$$\mathrm{a}_{\mathrm{k}} −\mathrm{a}_{\mathrm{k}−\mathrm{1}} =\mathrm{a}_{\mathrm{k}−\mathrm{1}} ^{\mathrm{2}} \\ $$$$\Leftrightarrow\underset{\mathrm{k}=\mathrm{3}} {\overset{\mathrm{n}+\mathrm{1}} {\sum}}\left(\mathrm{a}_{\mathrm{k}} −\mathrm{a}_{\mathrm{k}−\mathrm{1}} \right)=\underset{\mathrm{k}=\mathrm{2}} {\overset{\mathrm{n}} {\sum}}\mathrm{a}_{\mathrm{k}} ^{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{a}_{\mathrm{n}+\mathrm{1}} −\mathrm{a}_{\mathrm{2}} =\underset{\mathrm{k}=\mathrm{2}} {\overset{\mathrm{n}} {\sum}}\left(\mathrm{a}_{\mathrm{k}} \right)^{\mathrm{2}} \\ $$$$\mathrm{E}=\sqrt{\mathrm{9}+\mathrm{4}\left(\mathrm{a}_{\mathrm{n}+\mathrm{1}} −\mathrm{a}_{\mathrm{2}} \right)}=\sqrt{\mathrm{9}+\mathrm{4}\left(\mathrm{a}_{\mathrm{n}+\mathrm{1}} −\mathrm{2}\right)} \\ $$$$=\sqrt{\mathrm{1}+\mathrm{4a}_{\mathrm{n}+\mathrm{1}} }=\sqrt{\mathrm{1}+\mathrm{4a}_{\mathrm{n}} \left(\mathrm{a}_{\mathrm{n}} +\mathrm{1}\right)}=\sqrt{\left(\mathrm{2a}_{\mathrm{n}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\mid\mathrm{2a}_{\mathrm{n}} +\mathrm{1}\mid=\mathrm{2a}_{\mathrm{n}} +\mathrm{1},\mathrm{a}_{\mathrm{n}} >\mathrm{0} \\ $$

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