Vincen wrote:The sum of the first 50 positive even integers is 2,550. What is the sum of the first 50 odd integers?

(A) 2,000
(B) 2,497
(C) 2,500
(D) 2,501
(E) 2,549

The sum of the first 50 positive EVEN integers is 2,550
So, 2 + 4 + 6 + . . . + 98 + 100 = 2550

What is the sum of the first 50 odd integers?
So, we want the value of 1 + 3 + 5 + . . . + 97 + 99

IMPORTANT: Notice that:
1 = 2 - 1
3 = 4 - 1
5 = 6 - 1
.
.
.
97 = 98 - 1
99 = 100 - 1

So, 1 + 3 + 5 + . . . + 97 + 99 = (2 - 1) + (4 - 1) + (6 - 1) + . . . + (98 - 1) + (100 - 1)
= [2 + 4 + 6 + . . . + 98 + 100 ] - 1 - 1 - 1 - 1 . . . . -1 - 1
= [2550 ] - 1 - 1 - 1 - 1 . . . . -1 - 1

ASIDE: How many 1s are we subtracting from 2550 ?
Well, there are 50 even integers from 2 to 100, so we are subtracting fifty 1's
So, we get:
1 + 3 + 5 + . . . + 97 + 99 = [2550 ] - 1 - 1 - 1 - 1 . . . . -1 - 1
= [2550 ] - 50
= 2500

Answer: C

Cheers,
Brent

Vincen wrote:The sum of the first 50 positive even integers is 2,550. What is the sum of the first 50 odd integers?

(A) 2,000
(B) 2,497
(C) 2,500
(D) 2,501
(E) 2,549

Note that we can also solve this question WITHOUT using the fact that the sum of the first 50 positive even integers is 2,550

We need only look for a pattern in the sum of ODD integers.
1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = 25
1 + 3 + 5 + 7 + 9 + 11 = 36

Notice that each sum is the square of an integer (e.g., 4 = 2², 9 = 3², 16 = 4², etc)
So, it looks like the sum of the first 50 integers will be the square of an integer.
Since 2500 is the only square of an integer, it is most likely the correct answer.

Alternatively, we might recognize that:
The sum of the first 1 ODD integer = 1²
The sum of the first 2 ODD integers = 2²
The sum of the first 3 ODD integers = 3²
The sum of the first 4 ODD integers = 4²
.
.
.
So, the sum of the first 50 ODD integers = 50² = 2500

Cheers,
Brent

Vincen wrote: ↑

Fri Sep 29, 2017 6:14 pm

The sum of the first 50 positive even integers is 2,550. What is the sum of the first 50 odd integers?

(A) 2,000
(B) 2,497
(C) 2,500
(D) 2,501
(E) 2,549

The OA is C.

What is the relation between the sum of the even and the sum of the odd integers? I already know that the answer should be an even number. So options (B), (D) and (E) are out.

Solution:

We can use the formula sum = average x quantity, and if we have an evenly spaced set (notice that a set of consecutive odd integers is an evenly spaced set), average = (first number + last number)/2.

Therefore, since the first odd positive integer = 1 and the last (or in this case, the 50th) odd positive integer = 99, we have average = (1 + 99)/2 = 50. Since quantity = 50, then sum = average x quantity = 50 x 50 = 2500.

Alternate Solution:

Let’s say we don’t know any formulas involving the sum of an evenly spaced set. Let’s just use the given in the problem: “The sum of the first 50 positive even integers is 2,550.” In other words, we are given the sum of 2, 4, 6, …, 100 and we are asked to find the sum of the first 50 positive odd integers, which is the sum of 1, 3, 5, …, 99. Now notice that 1 is 1 less than 2, 3 is 1 less than 4, 5 is 1 less than 6, …, and 99 is 1 less than 100. In other words, each number in the set of the first 50 positive odd integers is 1 less than its counterpart in the set of the first 50 positive even integers. Since there are 50 numbers in each set, then the sum of the first 50 positive odd integers should be 50 less than the sum of the first 50 positive even integers. That is, the sum of the first 50 positive odd integers = 2550 - 50 = 2500.

Answer: C