Find $a_{2017}$: Sequence Challenge

[lfdahl]

Find $a_{2017}$, if $a_1 = 1$, and $$\frac{a_n}{n+1}=\frac{\sum_{i=1}^{n-1}a_i}{n-1}.$$

 

[Albert1]

Find $a_{2017}$, if $a_1 = 1$, and $$\frac{a_n}{n+1}=\frac{\sum_{i=1}^{n-1}a_i}{n-1}---(1).$$

Spoiler

$a_{2017}=1009\times 2^{2016}$ correct ?

 

[lfdahl]

Spoiler

$a_{2017}=1009\times 2^{2016}$ correct ?

Spoiler

Yes, that´s correct!

 

[Albert1]

my solution:

Spoiler

from(1) we have:
$a_1=1,a_2=3,a_3=8,a_4=20,a_5=48,------$
so we may set :$a_n=2a_{n-1}+2^{n-2}---(2)$
or $a_n-a_{n-1}=a_{n-1}+2^{n-2}--(3)$
and $S_{n-1}=a_n(\dfrac{n-1}{n+1})---(4)$
so $$a_{2017}-{\sum_{i=1}^{2016}a_i}=2^{2016}$$
or $a_{2017}-S_{2016}=2^{2016}$
from $(3)(4)$$a_{2017}=1009\times 2^{2016}$

 

[lfdahl]

my solution:

Spoiler

from(1) we have:
$a_1=1,a_2=3,a_3=8,a_4=20,a_5=48,------$
so we may set :$a_n=2a_{n-1}+2^{n-2}---(2)$
or $a_n-a_{n-1}=a_{n-1}+2^{n-2}--(3)$
and $S_{n-1}=a_n(\dfrac{n-1}{n+1})---(4)$
so $$a_{2017}-{\sum_{i=1}^{2016}a_i}=2^{2016}$$
or $a_{2017}-S_{2016}=2^{2016}$
from $(3)(4)$$a_{2017}=1009\times 2^{2016}$

Good job, Albert! Thankyou for your participation!