The sequence defined by the recurrence relation $$\frac{a_n}{n+1}=\frac{\sum_{i=1}^{n-1}a_i}{n-1}$$ with the initial condition $a_1 = 1$ leads to the calculation of $a_{2017}$. The discussion highlights the importance of understanding the relationship between the terms of the sequence and the summation of previous terms. Participants confirmed the validity of the approach to derive the specific term $a_{2017}$ through iterative calculations based on the established formula.
PREREQUISITESMathematics enthusiasts, students studying sequences, and educators looking for examples of recurrence relations in problem-solving contexts.
lfdahl said:Find $a_{2017}$, if $a_1 = 1$, and $$\frac{a_n}{n+1}=\frac{\sum_{i=1}^{n-1}a_i}{n-1}---(1).$$
Albert said:$a_{2017}=1009\times 2^{2016}$ correct ?
Albert said:my solution:
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}$