How do I derive the formula for sum of n consecutive natural numbers
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How do I derive the formula for sum of n consecutive natural numbers

[From: ] [author: ] [Date: 11-09-18] [Hit: ]
.qed-1 + 2 + 3 + ... + (n-1) + n= (1/2)* [ 1 + 2 + 3 + ........

So we need to find three coefficients, so we will need three points:

Let's choose the first three (although ANY three will work):

n = 1 --> S(1) = 1
n = 2 --> S(2) = 1 + 2 = 3
n = 3 --> S(3) = 1 + 2 + 3 = 6

So we have:

S(1) = a * 1² + b * 1 + c = 1
S(2) = a * 2² + b * 2 + c = 3
S(3) = a * 3² + b * 3 + c = 6
--> which yields this linear equation

a + b + c = 1
4a + 2b + c = 3
9a + 3b + c = 6

let's eliminate c by subtracting it from the first:

a + b + c = 1
3a + b + 0 = 2
8a + 2b + 0 = 5

Now let's solve the bottom two:

3a + b = 2
8a + 2b = 5
--> multiply first equation by 2

6a + 2b = 4
--> subtract from second

2a + 0 = 1
-->

a = 1/2

Go back and find b

6 * 1/2 + 2b = 4
-->

b = (4 - 3) / 2 = 1/2

Now find c:

1/2 + 1/2 + c = 1
-->

c = 1 - 1 = 0

So we have:

a = b = 1/2 and c = 0:

S(n) = 1/2 * n² + 1/2 * n + 0
--> or we can factor out 1/2 * n
-->

S(n) = n/2 * (n + 1) = n(n + 1) / 2...qed

-
1 + 2 + 3 + ... + (n-1) + n

= (1/2)* [
1 + 2 + 3 + ... + (n-1) + n +
1 + 2 + 3 + ... + (n-1) + n ] ..............( repeat the sum twice )

= (1/2)* [
1 + 2 + 3 + ... + (n-1) + n +
n + (n-1) + .....+ 3 + 2 + 1 ] ..............( reverse the second part )

= (1/2)* [ (1+n) + (2+(n-1)) + (3+(n-2)) + ... + (n+1) ]

= (1/2)* [ (n+1) + (n+1) + ... + (n+1) ] ...............( there are n (n+1)'s. )

= (1/2)* n* (n+1)

= n(n+1)/2

-
I've came across a step by step approach at http://www.exposix.com/members/answers/H… , where you can understand the terms used, as they aren't mathematical terms :)
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