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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