Sum of a Math Series

In our last discussion, we had the following formula:

Sum of a series of numbers = 1/2 of [ N x (F + L) ]

We used it to find the sum of consecutive numbers.

Suppose, we have the following series:

1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Is our formula applicable? We have ten numbers (N = 10). First number in the series is 1 (F = 1) and the last number is 19 (L = 19).


Using our formula we now have:

Sum of the series = 1/2 of [ 10 x (1 + 19) ]

Sum of the series = 1/2 x [200]

Sum of the series = 100 which is the correct answer.

The same formula applies not only to a series of consecutive numbers but also to other math series with constant difference such as 1, 3, 5, 7, 9, 11, 13, 15, 17, 19.

Math is fun

Math is fun. I remember as a young boy, I started arithmetic with counting.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20

Then continued with addition:

1 + 1 = 2
2 + 1 = 3
3 + 1 = 4
4 + 1 = 5

What if we try to sum up all the numbers? So for numbers 1, 2, 3, 4, 5, we get

Sum = 1 + 2 + 3 + 4 + 5 = 15

Using some addition property (called Commutative Property for Addition), we can re-arrange the numbers and get

Sum = 5 + 4 + 3 + 2 + 1
Sum = 1 + 2 + 3 + 4 + 5

Adding these two we get

Sum = 5 + 4 + 3 + 2 + 1
Sum = 1 + 2 + 3 + 4 + 5
============================
Sum + Sum = 6 + 6 + 6 + 6 + 6
2 times Sum = 5 times 6

So,

2 x Sum = 5 x 6

Note that 5 corresponds to the no. of consecutive numbers that were added.

Note also that 6 = sum of first and last number in the series
6 = 1 + 5


If we continue,
2 x Sum = 5 x (First Number + Last Number)
2 x Sum = (No. of consecutive numbers added) x (First Number + Last Number)

If we let N = no. of consecutive numbers that were added
F = First no. in the series
L = Last no. in the series

2 x Sum = N x (F + L)

Since we are interested in the Sum, we will get HALF of 2 x Sum:


Half of (2 x Sum) = Half of [ N x (F + L) ]

Or

Sum = 1/2 of [ N x (F + L) ]

You can use this formula to compute also the sum: 1 + 2 + 3 + . . . + 20

Since we are adding 20 consecutive numbers, N = 20.

The first number in the series is 1, so F = 1.

The last number in the series is 20, so L = 20.

So, Sum of 1 to 20 = 1/2 of [ N x (F + L) ]

Replacing actual values for N, F and L, we get

Sum of 1 to 20 = 1/2 of [ 20 x (1 + 20) ]

Sum of 1 to 20 = 1/2 of [ 20 x (21) ] = 210

Remember our formula:
Sum of a series of numbers = 1/2 of [ N x (F + L) ]