Kaori writes a sequence with the property that after the first two terms in the sequence, each term is equal to one more than the term before it, minus the term before that. In other words, \(t_n=1+t_{n-1}-t_{n-2}\), for \(n \geq 3\), where \(t_n\) denotes the \(n^\textrm{th}\) term in the sequence.
The first term in Kaori’s sequence is \(x\) and the second term is \(y\), where \(x\) and \(y\) are real numbers. That is, \(t_1 = x\) and \(t_2 = y\). Determine the sum of the first \(2021\) terms in her sequence, as an expression in terms of \(x\) and \(y\).
We are given that \(t_1 = x\), \(t_2 = y\), and \(t_n=1+t_{n-1}-t_{n-2}\), for \(n \geq 3\).
Let’s use the equation \(t_n=1+t_{n-1}-t_{n-2}\) for \(n\geq 3\). \[\begin{aligned}
t_3 & = 1 + t_2 - t_1 = 1 + y - x\\
t_4&=1+t_3-t_2=1+(1+y-x)-y=2-x\\
t_5&=1+t_4-t_3=1+(2-x)-(1+y-x)=2-y\\
t_6&=1+t_5-t_4=1+(2-y)-(2-x)=1-y+x\\
t_7&=1+t_6-t_5=1+(1-y+x)-(2-y)=x\\
t_8&=1+t_7-t_6=1+x-(1-y+x)=y\end{aligned}\] Since \(t_7=t_1\) and \(t_8=t_2\), and each term in the sequence depends only on the previous two terms, it follows that the sequence repeats every six terms.
The sum of the first six terms in the sequence is equal to \(x+y + (1+y-x)+(2-x)+(2-y)+(1-y+x)=6.\)
It follows that the sum of each successive group of six terms is also equal to \(6\).
We note that \(2022=6 \times 337\), so the \(2022^\textrm{nd}\) term of the sequence is the end of a group of six terms. Thus, the sum of the first \(2022\) terms in the sequence is equal to \(6 \times 337=2022\). It also follows that \(t_{2022}=t_6=1-y+x\).
Since the sum of the first \(2021\) terms is equal to the sum of the first \(2022\) terms minus the \(2022^\textrm{nd}\) term, we know that the sum of the first \(2021\) terms of the sequence is equal to \(2022 - (1-y+x)=2021+y-x\).