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) = ( (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) = ( (1 + 101)/(2) ).51 = 2601`