Problem E and Solution

Odd Sum

A sequence consists of \(2022\) terms. Each term after the first term is \(1\) greater than the previous term. The sum of the \(2022\) terms is \(31\,341\).

Determine the sum of the terms in the odd-numbered positions. That is, determine the sum of every second term starting with the first term and ending with the second last term.

Note:

In solving the above problem, it may be helpful to use the fact that the sum of the first \(n\) positive integers is equal to \(\tfrac{n(n+1)}{2}\). That is, \[1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}\]

**Solution 1**

Let \(t_n\) denote the \(n^{\text{th}}\) term in the sequence.

Let \(S_O\) represent the sum of the terms in the odd-numbered positions. That is, \[S_O=t_1+t_3+t_5+\cdots+t_{2021}\]

Let \(S_E\) represent the sum of the terms in the even-numbered positions. That is, \[S_E=t_2+t_4+t_6+\cdots+t_{2022}\] Since there are \(2022\) terms, and half of the terms of the sequence are in even-numbered positions and half are in odd-numbered positions, there are \(1011\) terms in \(S_O\) and \(1011\) terms in \(S_E\).

Let \(S\) represent the sum of the \(2022\) terms. That is, \[S=S_O+S_E=31\,341\tag{1}\]

Since each term after the first term is \(1\) greater than the term before, \[\begin{aligned} t_2&=t_1+1\\ t_4&=t_3+1\\ t_6&=t_5+1\end{aligned}\] and so on, until \[\begin{aligned} t_{2022}&=t_{2021}+1\end{aligned}\] Now, \[\begin{aligned} S_E&=t_2+t_4+t_6+\cdots+t_{2016}+t_{2022}\\ &=(t_1+1)+(t_3+1)+(t_5+1)+\cdots+(t_{2019}+1)+(t_{2021}+1)\\ &=(t_1+t_3+t_5+\cdots+t_{2019}+t_{2021})+ 1011\\ &=S_O+1011\end{aligned}\]

Substituting \(S_E = S_O+1011\) into equation \((1)\), \[\begin{aligned} S_O+S_E&=31\,341\\ S_O+S_O+1011&=31\,341\\ 2S_O&=31\,341-1011\\ 2S_O&=30\,330\\ S_O&=15\,165\end{aligned}\]

Therefore, the sum of the terms in the odd-numbered positions is \(15\,165\).

Notice that this solution did not need the formula given in the note after the problem.

**Solution 2**

Let \(t_1\) represent the first term in the sequence. Every term in the sequence can be written in terms of \(t_1\). The second term is \(1\) more than the first term, the third term is \(2\) more than the first term, the fourth term is \(3\) more than the first term, and so on. Thus, \[\begin{aligned} t_1+t_2+t_3+t_4+\cdots+t_{2021}+t_{2022}&=31\,341\\ t_1+(t_1+1)+(t_1+2)+(t_1+3)+\cdots+(t_1+2020)+(t_1+2021)&=31\,341\\ 2022t_1+(1+2+3+\cdots+2020+2021)&=31\,341\end{aligned}\] Using the formula for the sum of the first \(n\) positive integers with \(n=2021\), \[2022t_1+\frac{2021(2022)}{2}=34\,341\] Dividing by \(2022\), \[\begin{aligned} t_1+\frac{2021}{2}&=\frac{31\,341}{2022}\\ t_1+1010.5&=15.5\\ t_1&=-995\end{aligned}\]

Since the first term in the sequence is \(-995\), we know that the original sum is \[\begin{aligned} t_1+t_2+t_3+t_4+&\cdots+t_{2020}+t_{2021}+t_{2022} \\ &=t_1+(t_1+1)+(t_1+2)+(t_1+3)+\cdots+(t_1+2020)+(t_1+2021)\\ &=-995-994-993-992-\cdots+1025+1026\end{aligned}\]

We are interested in the sum of the terms in the odd-numbered positions. That is, we’re interested in the sum \[-995-993-991-\cdots+991 + 993 + 995 + 997 + 999+ \cdots + 1023+1025\]

From this point, we will present two different methods for determining this sum.

*Method 1:*Notice that this sum includes all of the odd integers from \(-995\) to \(995\), inclusive. This sum is \(0\). Thus, \[\begin{aligned} -995-993-991-\cdots+991 + 993 +& 995 + 997 + 999 + \cdots + 1023+1025\\ &= 0 + 997 + 999 + \cdots + 1023 + 1025\\ &= 997(15) + 2 + 4 + \cdots + 28\\ &= 14\,955 + 2(1 + 2 + \cdots +14)\\ &= 14\,955 + 2\left[\frac{14(15)}{2}\right]\\ &= 14\,955 +210\\ &= 15\,165\end{aligned}\]

*Method 2:*This is an arithmetic series with \(n=1011\) terms, first term \(t_1 = -995\), and last term \(t_n = 1025\).

Then, using the formula \(S_n = {n}\left[ \dfrac{t_1 + t_n}{2} \right]\) for the sum of a series, \[\begin{aligned} -995-993-991-989-\cdots+1023+1025&=1011\left[\frac{-995+1025}{2}\right]\\ &=1011(15)\\ &=15\,165\end{aligned}\]

Therefore, the sum of the terms in the odd-numbered positions is \(15\,165\).