Suppose t(n) is the nth term.
Then t(n) = -1 + 2n (for n = 1, 2, 3, ... 100)
Now
t(1) + t(100) = (-1 + 2*1) + (-1 +2*100) = 1 + 199 = 200
t(2) + t(99) = (-1 + 2*2) + (-1 +2*99) = 3 + 197 = 200
there are 50 such pairs.
So the grand total is 50*200 = 10000
They are 2n+2
It is 2500.
The sum of the first 230 positive, odd integers is 230 squared or 52,900.
The sum of the first 5,000 odd, positive integers is 5,000 squared or 25,000,000.
The first 100 positive odd integers are the numbers that can be expressed in the form of (2n - 1), where (n) is a positive integer. They start from 1 and go up to 199. Specifically, the sequence is: 1, 3, 5, 7, ..., up to 199. Each subsequent odd integer is obtained by adding 2 to the previous one.
They are 2n+2
The sum of the first eleven positive odd integers is 121.
The sum of the first 2,006 positive, odd integers is 4,024,036.
The first odd positive integers are "1" and "3" which the sum is 4.
find the two consecutive odd integers with a sum of 152
It is 2500.
The sum of the first 230 positive, odd integers is 230 squared or 52,900.
The sum of the first 5,000 odd, positive integers is 5,000 squared or 25,000,000.
The first 100 positive odd integers are the numbers that can be expressed in the form of (2n - 1), where (n) is a positive integer. They start from 1 and go up to 199. Specifically, the sequence is: 1, 3, 5, 7, ..., up to 199. Each subsequent odd integer is obtained by adding 2 to the previous one.
The sum of the squares of the first 1000 positive odd integers (from 12 to 19992) is 1333333000.
The sum of the first positive odd integers less than 101 is 10,000.
Yes. The sum of the first 5,000 odd, positive integers is 25,000,000 (25 million).