Sum of even integers !

Question

For any positive integers n, the sum of the first n positive integers equals (n(n+1))/2. What is the sum of all the even integers between 99 and 301 ?
a) 10,100
b) 20,200
c) 22,650
d) 40,200
e) 45,150

Answer: B

Thanks in advance.

Brent@GMATPrepNow · Answer

Uva@90 wrote:For any positive integers n, the sum of the first n positive integers equals (n(n+1))/2. What is the sum of all the even integers between 99 and 301 ?
a) 10,100
b) 20,200
c) 22,650
d) 40,200
e) 45,150

Answer: B

Thanks in advance.

Here's one approach.

We want 100+102+104+....298+300
This equals 2(50+51+52+...+149+150)
From here, a quick way is to evaluate this is to first recognize that there are 101 integers from 50 to 150 inclusive (150 - 50 + 1 = 101)

To evaluate 2(50+51+52+...+149+150), let's add values in pairs:

....50 + 51 + 52 +...+ 149 + 150
+150+ 149+ 148+...+ 51 + 50
...200+ 200+ 200+...+ 200 + 200

How many 200's do we have in the new sum? There are 101 altogether.
101 x 200 = 20,200 = B

Cheers,
Brent

Last edited by Brent@GMATPrepNow on Fri Jan 26, 2018 3:24 pm, edited 3 times in total.

Brent Hanneson - Creator of GMATPrepNow.com

Brent@GMATPrepNow · Answer

Uva@90 wrote:For any positive integers n, the sum of the first n positive integers equals (n(n+1))/2. What is the sum of all the even integers between 99 and 301 ?
a) 10,100
b) 20,200
c) 22,650
d) 40,200
e) 45,150

Answer: B

Thanks in advance.

Here's one approach.

We want 100+102+104+....298+300
This equals 2(50+51+52+...+149+150)
From here, a quick way is to evaluate this is to first recognize that there are 101 integers from 50 to 150 inclusive (150 - 50 + 1 = 101)

To evaluate 2(50+51+52+...+149+150), let's add values in pairs:

....50 + 51 + 52 +...+ 149 + 150
+150+ 149+ 148+...+ 51 + 50
...200+ 200+ 200+...+ 200 + 200

How many 200's do we have in the new sum? There are 101 altogether.
101 x 200 = 20,200 = B

Cheers,
Brent

Last edited by Brent@GMATPrepNow on Fri Jan 26, 2018 3:24 pm, edited 3 times in total.

Brent Hanneson - Creator of GMATPrepNow.com

Brent@GMATPrepNow · Answer

Uva@90 wrote:For any positive integers n, the sum of the first n positive integers equals (n(n+1))/2. What is the sum of all the even integers between 99 and 301 ?
a) 10,100
b) 20,200
c) 22,650
d) 40,200
e) 45,150

Answer: B

Thanks in advance.

Here's one approach.

We want 100+102+104+....298+300
This equals 2(50+51+52+...+149+150)
From here, a quick way is to evaluate this is to first recognize that there are 101 integers from 50 to 150 inclusive (150 - 50 + 1 = 101)

To evaluate 2(50+51+52+...+149+150), let's add values in pairs:

....50 + 51 + 52 +...+ 149 + 150
+150+ 149+ 148+...+ 51 + 50
...200+ 200+ 200+...+ 200 + 200

How many 200's do we have in the new sum? There are 101 altogether.
101 x 200 = 20,200 = B

Cheers,
Brent

Last edited by Brent@GMATPrepNow on Fri Jan 26, 2018 3:24 pm, edited 3 times in total.

Brent Hanneson - Creator of GMATPrepNow.com

Uva@90 · Answer

Brent@GMATPrepNow wrote:Uva@90 wrote:For any positive integers n, the sum of the first n positive integers equals (n(n+1))/2. What is the sum of all the even integers between 99 and 301 ? a) 10,100 b) 20,200 c) 22,650 d) 40,200 e) 45,150 Answer: B Thanks in advance. Approach #3: Take 100+102+104+ ...+298+300 and factor out the 2 to get 2(50+51+52+...+149+150) From here, we'll evaluate the sum 50+51+52+...+149+150, and then double it. Important: notice that 50+51+.....149+150 = (sum of 1 to 150) - (sum of 1 to 49) Now we use the given formula: sum of 1 to 150 = 150(151)/2 = 11,325 sum of 1 to 49 = 49(50)/2 = 1,225 So, sum of 50 to 150 = 11,325 - 1,225 = 10,100 So, 2(50+51+52+...+149+150) = 2(10,100) = [spoiler]20,200 = B[/spoiler] Cheers, Brent Brent : Thanks a Ton. You took your time to explaining me all the possible ways to solve the problem. I thank you once again. Regards, Uva.

send2dar · Answer

Thanks Brent.. I felt approach 2 is the quickest... I liked approach 1 too.. Regards --------- "Trying for 750+"

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Sum of even integers !

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