# Problem of the Week Problem E and Solution Odd Sum

## Problem

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

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 $$15\,165$$.