** Sum of Increasing Series**

**The Sum From 1 to N - Increasing by 1**

It is well known that sum from 1 to N, which each term increasing by 1 is:

1 + 2 + 3 + 4 + ... + N = ∑ x from x = 1 to N = (N + 1) * N / 2

The derivation is fairly easy. Let S be the sum:

S = 1 + 2 + 3 + ... + N-1 + N

Add S to both sides:

2 * S = 1 + 2 + 3 + ... + N-1 + N + 1 + 2 + 3 ... + N-1 + N

Thanks to the commutative property of addition, we can arrange terms, and with a clever and creative way of arranging terms:

2 * S = ( 1 + N ) + ( 2 + N - 1 ) + ( 3 + N - 2 ) + ... + ( N - 1 + 2 ) + ( N + 1 )

Note that:

2 * S =

1 + 2 + 3 + ... + N-1 + N +

N + N-1 + N-2 + .. + 2 + 1

Written this way, there are N "pairs". Hence:

2 * S = ( 1 + N ) + ( N + 1 ) + ( N +1 ) + ... + ( N + 1 ) + ( N + 1 )

2 * S = ( N + 1 ) * N

Solving for S:

S = ( N + 1 ) * N / 2

**The Sum From 1 to N - Increasing by 2**

Now lets consider the sum:

1 + 3 + 5 + 7 + ...

Fun fact: Adding odd numbers in this fashion will always total perfect squares.

1 + 3 = 4 = 2^2

1 + 3 + 5 = 9 = 3^2

1 + 3 + 5 + 7 = 16 = 4^2

1 + 3 + 5 + 7 + 9 = 25 = 5^2

... and so on

Let's define S as the sum:

S = 1 + 3 + 5 + ... + N-4 + N-2 + N

Note that

1 = 2 * 0 + 1

3 = 2 * 1 + 1

5 = 2 * 2 + 1

and so on...

For integer q,

N = 2 * q + 1

Then:

S = 1 + 3 + 5 + ... + ( 2*(q-2) + 1 ) + ( 2*(q-1) + 1) + ( 2*q + 1 )

Using the same strategy as last time:

2 * S = 1 + 3 + 5 + ... + ( 2*(q-2) + 1 ) + ( 2*(q-1) + 1) + ( 2*q + 1 )

+ 1 + 3 + 5 + ... + ( 2*(q-2) + 1 ) + ( 2*(q-1) + 1) + ( 2*q + 1 )

2 * S =

1 + 3 + 5 + ... + (2*(q-2) + 1) + (2*(q-1) + 1) + (2*q + 1) +

(2*q + 1) + (2*(q-1) + 1) + (2*(q-2) + 1) + ... + 5 + 3 + 1

Combine each pair as such:

2 * S = (2*q + 2) + (2*q + 2) + (2*q + 2) + ... + (2*q + 2) + (2*q + 2) + (2*q + 2)

Since q starts at q = 0, there are q+1 "pairs".

2 * S = (q + 1) * (2*q + 2)

S = (q + 1) * (2*q + 2) / 2

To show that the sum is a perfect square:

S = (q + 1) * (2*q + 2) / 2

= (2*q^2 + 2*q + 2*q + 2) / 2

= (2*q^2 + 4*q + 2) / 2

= (q^2 + 2*q + 1)

= (q + 1)^2

Example:

Calculate 7^2 (albeit the "longer" way):

7 = q+1; q = 6:

S = (6 + 1) * (2*6 + 2) / 2 = 7 * 14 / 2 = 7 * 7 = 49

**The Sum From 1 to N - Increasing by 3**

Now add the series:

S = 1 + 4 + 7 + 10 + ... + N - 3 + N

where each term increases by 3. Letting q be a positive integer and noting that

1 = 3*0 + 1

4 = 3*1 + 1

7 = 3*2 + 1

10 = 3*3 + 1

and so on...

Then:

S = (3*0 + 1) + (3*1 + 1) + (3*2 + 1) + .... + (3(q-1) + 1) + (3q + 1)

2* S

= (3*0 + 1) + (3*1 + 1) + (3*2 + 1) + ... + (3(q-1) + 1) + (3q + 1)

+ (3*0 + 1) + (3*1 + 1) + (3*2 + 1) + ... + (3(q-1) + 1) + (3q + 1)

2 * S

= (3*0 + 1) + (3*1 + 1) + (3*2 + 1) + ... + (3(q-1) + 1) + (3q + 1)

+ (3*q + 1) + (3(q-1) + 1) +(3(q-2) + 1) + ... + (3*1 + 1) + (3*0 + 1)

There are q+1 pairs.

2 * S

= (3*q + 2) + (3*q + 2) + (3*q + 2) + ... + (3*q + 2) + (3*q + 2)

2 *S = (q + 1) * (3*q + 2)

S = (q + 1) * (3*q + 2) / 2

Example:

q = 5 (terms from 0 to 5)

S = (5 + 1) * (3*5 + 2) / 2 = 6 * 17 / 2 = 51

1 + 4 + 7 + 10 + 13 + 16 = 51

**The Sum From A for q terms, increasing by D**

Let A ≥ 0, D > 0 and q be a positive integer, and A is a starting term, let the sum S be:

S = A + (A + D) + (A + 2*D) + (A + 3*D) + ... + (A + (q-2)*D) + (A + (q-1)*D) + (A + q*D)

Since we start at q = 0, there are q+1 terms.

2*S =

A + (A + D) + (A + 2*D) + (A + 3*D) + ... + (A + (q-2)*D) + (A + (q-1)*D) + (A + q*D) +

A + (A + D) + (A + 2*D) + (A + 3*D) + ... + (A + (q-2)*D) + (A + (q-1)*D) + (A + q*D)

2*S =

A + (A + D) + (A + 2*D) + ... + (A + (q-2)*D) + (A + (q-1)*D) + (A + q*D) +

(A + q*D) + (A + (q-1)*D) + (A + (q-2)*D) + ... + (A + 2*D) + (A + D) + A

2*S = (2*A + q*D) + (2*A + q*D) + (2*A + q*D) + ... + (2*A + q*D) + (2*A + q*D) + (2*A + q*D)

2*S = (q + 1) * (2*A + q*D)

S = (q + 1) * (2*A + q*D) / 2

Example:

A = 10, D = 7; q = 4 (5 terms, increase by 7, initial term is 10)

(4 + 1) * (2*10 + 4*7) / 2 = 5 * (20 + 28) / 2 = 120

Note: 10 + 17 + 24 + 31 + 38 = 120

Source:

Knott, Dr. Ron "Proving that 1+2+3+...+n is n(n+1)/2" February 12, 2003. Accessed December 6, 2020. http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/runsums/triNbProof.html

Eddie

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