Let-a-sequence-a-n-satisfies-a-1-1-na-n-n-2-k-1-n-1-a-k-n-gt-2-Find-the-value-of-a-2021-

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

Let a sequence {a_{n}} satisfies

{\begin{cases} a_{1} = 1 \\ {n a}_{n} = n + 2 \underset{k = 1}{\sum^{n - 1}} a_{k}, n > 2 \end{cases}

Find the value of a_{2021} .

Commented by JDamian last updated on 15/Jan/21

a_{2} i s n o t d e f i n e d, {i s n}^{'} t i t ?

Answered by Olaf last updated on 15/Jan/21

2 a_{2} = 2 + 2 a_{1} = 4 \Rightarrow a_{2} = 2

3 a_{3} = 3 + 2 (a_{2} + a_{1}) = 9 \Rightarrow a_{3} = 3

4 a_{4} = 4 + 2 (a_{3} + a_{2} + a_{1}) = 16 \Rightarrow a_{4} = 4

5 a_{5} = 5 + 2 (a_{4} + a_{3} + a_{2} + a_{1}) = 25 \Rightarrow 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 :

{n a}_{n} = n + 2 \underset{k = 1}{\sum^{n - 1}} a_{k} = n + 2 \underset{k = 1}{\sum^{n - 1}} k

{n a}_{n} = n + 2 \frac{(n - 1) n}{2} = n^{2}

\Rightarrow a_{n} = n

a_{2021} = 2021

Commented by JDamian last updated on 16/Jan/21

b u t t h a t m e t h o d i s f o r n > 2, n o t f o r n = 2

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