Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 129413 by ZiYangLee last updated on 15/Jan/21

Let a sequence {a_n } satisfies { (( a_1 =1)),((na_n =n+2Σ_(k=1) ^(n−1) a_k , n>2)) :} Find the value of a_(2021) .

$$\mathrm{Let}\:\mathrm{a}\:\mathrm{sequence}\:\left\{{a}_{{n}} \right\}\:\mathrm{satisfies} \\ $$ $$\:\:\:\:\begin{cases}{\:\:\:\:\:\:\:\:\:\:\:\:{a}_{\mathrm{1}} =\mathrm{1}}\\{{na}_{{n}} ={n}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{a}_{{k}} ,\:{n}>\mathrm{2}}\end{cases} \\ $$ $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{a}_{\mathrm{2021}} . \\ $$

Commented byJDamian last updated on 15/Jan/21

a_2 is not defined, isn′t it?

$${a}_{\mathrm{2}} \:{is}\:{not}\:{defined},\:{isn}'{t}\:{it}? \\ $$

Answered by Olaf last updated on 15/Jan/21

2a_2 = 2+2a_1 = 4 ⇒ a_2 = 2 3a_3 = 3+2(a_2 +a_1 ) = 9 ⇒ a_3 = 3 4a_4 = 4+2(a_3 +a_2 +a_1 ) = 16 ⇒ a_4 = 4 5a_5 = 5+2(a_4 +a_3 +a_2 +a_1 ) = 25 ⇒ a_5 = 5 a_n = n ? It′s true for n = 1 : a_1 = 1 Suppose a_k = k, k ≤ n−1, then by induction : na_n = n+2Σ_(k=1) ^(n−1) a_k = n+2Σ_(k=1) ^(n−1) k na_n = n+2(((n−1)n)/2) = n^2 ⇒ a_n = n a_(2021) = 2021

$$\mathrm{2}{a}_{\mathrm{2}} \:=\:\mathrm{2}+\mathrm{2}{a}_{\mathrm{1}} \:=\:\mathrm{4}\:\Rightarrow\:{a}_{\mathrm{2}} \:=\:\mathrm{2} \\ $$ $$\mathrm{3}{a}_{\mathrm{3}} \:=\:\mathrm{3}+\mathrm{2}\left({a}_{\mathrm{2}} +{a}_{\mathrm{1}} \right)\:=\:\mathrm{9}\:\Rightarrow\:{a}_{\mathrm{3}} \:=\:\mathrm{3} \\ $$ $$\mathrm{4}{a}_{\mathrm{4}} \:=\:\mathrm{4}+\mathrm{2}\left({a}_{\mathrm{3}} +{a}_{\mathrm{2}} +{a}_{\mathrm{1}} \right)\:=\:\mathrm{16}\:\Rightarrow\:{a}_{\mathrm{4}} \:=\:\mathrm{4} \\ $$ $$\mathrm{5}{a}_{\mathrm{5}} \:=\:\mathrm{5}+\mathrm{2}\left({a}_{\mathrm{4}} +{a}_{\mathrm{3}} +{a}_{\mathrm{2}} +{a}_{\mathrm{1}} \right)\:=\:\mathrm{25}\:\Rightarrow\:{a}_{\mathrm{5}} \:=\:\mathrm{5} \\ $$ $${a}_{{n}} \:=\:{n}\:? \\ $$ $$\mathrm{It}'\mathrm{s}\:\mathrm{true}\:\mathrm{for}\:{n}\:=\:\mathrm{1}\::\:{a}_{\mathrm{1}} \:=\:\mathrm{1} \\ $$ $$\mathrm{Suppose}\:\mathrm{a}_{{k}} \:=\:{k},\:{k}\:\leqslant\:{n}−\mathrm{1},\:\mathrm{then}\:\mathrm{by}\:\mathrm{induction}\:: \\ $$ $${na}_{{n}} \:=\:{n}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{a}_{{k}} \:=\:{n}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{k} \\ $$ $${na}_{{n}} \:=\:{n}+\mathrm{2}\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}}\:=\:{n}^{\mathrm{2}} \\ $$ $$\Rightarrow\:{a}_{{n}} \:=\:{n} \\ $$ $$ \\ $$ $${a}_{\mathrm{2021}} \:=\:\mathrm{2021} \\ $$

Commented byJDamian last updated on 16/Jan/21

but that method is for n>2, not for n=2

$${but}\:{that}\:{method}\:{is}\:{for}\:{n}>\mathrm{2},\:{not}\:{for}\:{n}=\mathrm{2} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com