The sum of the first 100 positive integers is 5050 . That is, $1+2+\cdots+99+100=5050$. What is the sum of the first 100 positive odd integers?
(A) 5050
(B) 10000
(C) 10050
(D) 10100
(E) 10150
The first positive odd integer is 1 , the second is $2(2)-1=3$, the third is $2(3)-1=5$, the fourth is $2(4)-1=7$.

That is, the one hundredth positive odd integer is $2(100)-1=199$ and the sum that we are being asked to determine is $1+3+5+\cdots+195+197+199$.

## Solution 1

Since each odd integer can be expressed as the sum of two consecutive integers, we rewrite $1+3+5+\cdots+195+197+199$ as $1+(1+2)+(2+3)+\cdots+(97+98)+(98+99)+(99+100)$.

Rearranging the terms in the previous sum, we get $(1+2+3+\cdots+98+99+100)+(1+2+3+\cdots+97+98+99)$.

The sum in the first set of brackets, $1+2+3+\cdots+98+99+100$ is equal to 5050 .

The sum in the second set of brackets, $1+2+3+\cdots+98+99$ is 100 less than 5050 or 4950 . Therefore, $1+3+5+\cdots+195+197+199=5050+4950=10000$.

Solution 2

Doubling each term on the left of the equality $1+2+3+\cdots+98+99+100=5050$, doubles the result on the right.

That is, $2+4+6+\cdots+196+198+200=10100$.

Subtracting 1 from each term on the left side of this equality gives

$(2-1)+(4-1)+(6-1)+\cdots+(196-1)+(198-1)+(200-1)$ or $1+3+5+\cdots+195+197+199$, the required sum.

Since there are 100 terms on the left side of the equality, then subtracting one 100 times reduces the left side of the equality by 100 .

Subtracting 100 from the right side, we get $10100-100=10000$, and so

$1+3+5+\cdots+195+197+199=10000$.

## Solution 3

It is also possible to determine the sum of the first 100 positive odd integers, $1+3+5+\cdots+195+197+199$, without using the result given in the question.

Reorganizing the terms into sums of pairs, first the largest number with the smallest, then the next largest number with the next smallest, and so on,

$1+3+5+\cdots+195+197+199=(1+199)+(3+197)+(5+195)+\cdots+(97+103)+(99+101)$.

The sum of the first pair, $1+199$ is 200 .

Each successive pair following $1+199$ includes a number that is 2 more than a number in the previous pair, and also a number that is 2 less than a number in the previous pair, so then the sum of every pair is also 200 .

The total number of terms in the sum is 100, and thus there are 50 pairs of numbers each of which has a sum of 200 .

Therefore, $1+3+5+\cdots+195+197+199=50 \times 200=10000$.

ANSwER: (B)