MATH SOLVE

4 months ago

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

Accepted Solution

A:

We call:

[tex] a_{n} [/tex] as the set of the first 51 consecutive odd positive integers, so:

[tex]a_{n} = \{1, 3, 5, 7, 9...\} [/tex]

Where:

[tex]a_{1} = 1[/tex]

[tex]a_{2} = 3 [/tex]

[tex]a_{3} = 5 [/tex]

[tex]a_{4} = 7 [/tex]

[tex]a_{5} = 9 [/tex]

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:

[tex]a_{n} = \{ a_{1} , a_{1} + d, a_{1} + d + d,..., a_{1} + (n-2)d+d\} [/tex]

Then:

[tex]a_{n} = a_{1} + (n-1)d[/tex]

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:

[tex] S_{k} = ( \frac{a_{1} + a_{k}}{2} ).k[/tex]

Therefore, we need to find:

[tex] a_{k} = a_{51} [/tex]

Given that [tex]a_{1} = 1[/tex], then:

[tex]a_{n} = a_{1} + (n-1)d = 1 + (n-1)(2) = 2n-1 [/tex]

Thus:

[tex]a_{k} = a_{51} = 2(51)-1 = 101[/tex]

Lastly:

[tex] S_{51} = ( \frac{1 + 101}{2} ).51 = 2601[/tex]

[tex] a_{n} [/tex] as the set of the first 51 consecutive odd positive integers, so:

[tex]a_{n} = \{1, 3, 5, 7, 9...\} [/tex]

Where:

[tex]a_{1} = 1[/tex]

[tex]a_{2} = 3 [/tex]

[tex]a_{3} = 5 [/tex]

[tex]a_{4} = 7 [/tex]

[tex]a_{5} = 9 [/tex]

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:

[tex]a_{n} = \{ a_{1} , a_{1} + d, a_{1} + d + d,..., a_{1} + (n-2)d+d\} [/tex]

Then:

[tex]a_{n} = a_{1} + (n-1)d[/tex]

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:

[tex] S_{k} = ( \frac{a_{1} + a_{k}}{2} ).k[/tex]

Therefore, we need to find:

[tex] a_{k} = a_{51} [/tex]

Given that [tex]a_{1} = 1[/tex], then:

[tex]a_{n} = a_{1} + (n-1)d = 1 + (n-1)(2) = 2n-1 [/tex]

Thus:

[tex]a_{k} = a_{51} = 2(51)-1 = 101[/tex]

Lastly:

[tex] S_{51} = ( \frac{1 + 101}{2} ).51 = 2601[/tex]