Arithmetic Series

A series means
the sum of the terms of a sequence.

S_n = a₁ + a₂ + a₃ + ... + a_n

For an arithmetic sequence,
S_n = [n/2][2a + (n - 1)d].

S_n: The sum from a₁ to a_n
a: First term, a₁
d: Common difference

If a₁ and a_n are given,
you can also use this formula.

S_n = [n/2][a₁ + a_n]

a = 3
d = 5
n = 20

Then S₂₀ = [20/2][2⋅3 + 19⋅5].

20/2 = 10

2⋅3 = 6
+19⋅5 = +95

6 + 95 = 101

10⋅101 = 1010

So S₂₀ = 1010.

To find the number of the terms, n,
first find a_n.

Write -1, 3, 7, 10.

The first term, a, is -1.

-1 + 4 = 3
3 + 4 = 7
7 + 4 = 11

So write, the d, +4
between the terms.

a = -1
d = 4

So a_n = -1 + (n - 1)⋅4.

+(n - 1)⋅4 = +4n - 4

Multiply a Monomial and a Polynomial

-1 + 4n - 4 = 4n - 5

So a_n = 4n - 5.

The last term, a_n, is 31.

a_n = 4n - 5

So 4n - 5 = 31.

Move -5 to the right side.

Then 4n = 36.

Divide both sides by 4.

Then n = 9.

a = -1
d = 4
n = 9

Then S₉ = [9/2][2⋅(-1) + 8⋅4].

You can also use the other formula.

n = 9
a₁ = -1
a_n = 31

Then S₉ = [9/2][-1 + 31].

2⋅(-1) = -2
+8⋅4 = +32

-2 + 32 = 30

Cancel the denominator 2
and reduce 30 to, 30/2, 15.

9⋅15 = 135

So 135 is the answer.

S_n is given.
And it says to find a_n.

In this case,
think of two cases:
1) n = 2, 3, 4, ... and 2) n = 1.

Case 1: n = 2, 3, 4, ...

S_n = a₁ + a₂ + a₃ + ... + a_{n - 1} + a_n
S_{n - 1} = a₁ + a₂ + a₃ + ... + a_{n - 1}

Subtract these two equations.
Then S_n - S_{n - 1} = a_n.

So set a_n = S_n - S_{n - 1}.

This is true
when n = 2, 3, 4, ... .

S_n = n² + 2n

So write n² + 2n.

Write -.

S_n = n² + 2n

So
S_{n - 1} = (n - 1)² + 2(n - 1).

So
S_n - S_{n - 1}
= n² + 2n - [(n - 1)² + 2(n - 1)].

(n - 1)² = n² - 2⋅n⋅1 + 1²

Square of a Difference: (a - b)²

+2(n - 1) = +2n - 2

-2⋅n⋅1 = -2n
+1² = +1

Cancel n².
Cancel +2n.

+1 - 2 = -1

-[-2n - 1] = 2n + 1

So a_n = 2n + 1
when n = 2, 3, 4, ... .

Next, see case 2: n = 1.

When n = 1,
a_n = a₁.

S₁ = a₁

So a₁ = S₁.

S_n = n² + 2n

So S₁ = 1² + 2⋅1.

1² = 1
+2⋅1 = +2

1 + 2 = 3

So a₁ = 3.

You got a₁ = 3.

Then see if a_n = 2n + 1,
the a_n from case 1,
is true for n = 1.

In other words,
see if a₁ from a_n = 2n + 1
is equal to 3.

a₁ = 2⋅1 + 1

2⋅1 = 2

2 + 1 = 3

So a₁ = 3.

This a₁ is from a_n = 2n + 1.

a₁ from S₁
is 3.

And a₁ from a_n = 2n + 1
is also 3.

a₁ from a_n
is equal to
a₁ from S₁.

Then a_n = 2n + 1
is true for n = 1.

(If a₁ from a_n
is not equal to
a₁ from S₁,

then a₁ = S₁
and a_n is not true for n = 1.)

Case 1:
a_n = 2n + 1
(n = 2, 3, 4, ...)

Case 2:
a_n = 2n + 1
(n = 1)

So a_n = 2n + 1.

So a_n = 2n + 1.