Question

1 + 3 + 5 + ... + 99 =?

Answer 1

Будет равняться 2500

Answer 2

To find the sum of an arithmetic series like the one you've presented (1 + 3 + 5 + ... + 99), you can use the formula for the sum of the first n terms of an arithmetic series:

Sum = (n/2) * (first term + last term)

In this case, the first term is 1, the last term is 99, and the common difference between each term is 2 (3 - 1 = 2, 5 - 3 = 2, and so on).

We can calculate the number of terms (n) using the formula for the nth term of an arithmetic series:

nth term = first term + (n - 1) * common difference

Since we want to find the last term, which is 99, we can set up the equation:

99 = 1 + (n - 1) * 2

Now, solve for n:

98 = 2 * (n - 1) n - 1 = 98 / 2 n - 1 = 49 n = 50

Now that we know there are 50 terms, we can calculate the sum:

Sum = (n/2) * (first term + last term) Sum = (50/2) * (1 + 99) Sum = 25 * 100 Sum = 2500

So, the sum of the series 1 + 3 + 5 + ... + 99 is 2500.