Question

What is the sum of the first 51 consecutive odd positive integers?

Answer 1

We call:

$a_{n}$ as the set of the first 51 consecutive odd positive integers, so:

$a_{n} = \{1, 3, 5, 7, 9...\}$

Where:
$a_{1} = 1$
$a_{2} = 3$
$a_{3} = 5$
$a_{4} = 7$
$a_{5} = 9$
and so on.

In mathematics, a sequence of numbers, such that the difference between two consecutive terms is constant, is called Arithmetic Progression, so:

3-1 = 2
5-3 = 2
7-5 = 2
9-7 = 2 and so on.

Then, the common difference is 2, thus:

$a_{n} = \{ a_{1} , a_{1} + d, a_{1} + d + d,..., a_{1} + (n-2)d+d\}$

Then:

$a_{n} = a_{1} + (n-1)d$

So, we need to find the sum of the members of the finite series, which is called arithmetic series:

There is a formula for arithmetic series, namely:

$S_{k} = ( \frac{a_{1} + a_{k}}{2} ).k$

Therefore, we need to find:
$a_{k} = a_{51}$  

Given that $a_{1} = 1$, then:

$a_{n} = a_{1} + (n-1)d = 1 + (n-1)(2) = 2n-1$

Thus:
$a_{k} = a_{51} = 2(51)-1 = 101$

Lastly:

$S_{51} = ( \frac{1 + 101}{2} ).51 = 2601$