# Coded Data Proofs: Mean & Standard Deviation, y=x/k + C

Coded Data Proofs (4):

Say that: y=x/k + C

And that: x={p, q} and y={p/k+C, q/k+C}

If this is the case: $\frac { \Sigma y }{ n } =\frac { \frac { p }{ k } +C+\left\{ \frac { q }{ k } +C \right\} }{ n } \\ \\ =\frac { \frac { p }{ k } +\frac { q }{ k } +nC }{ n } \\ \\ =\frac { \frac { 1 }{ k } \left( p+q \right) +nC }{ n } \\ \\ =\frac { \frac { 1 }{ k } \left( p+q \right) }{ n } +\frac { nC }{ n } \\ \\ =\frac { 1 }{ k } \cdot \frac { \Sigma x }{ n } +C$

And also: $\frac { \Sigma { y }^{ 2 } }{ n } =\frac { { \left( \frac { p }{ k } +C \right) }^{ 2 }+{ \left( \frac { q }{ k } +C \right) }^{ 2 } }{ n } \\ \\ =\frac { \left( \frac { p }{ k } +C \right) \left( \frac { p }{ k } +C \right) +\left( \frac { q }{ k } +C \right) \left( \frac { q }{ k } +C \right) }{ n } \\ \\ =\frac { \frac { { p }^{ 2 } }{ { k }^{ 2 } } +n\cdot \frac { p }{ k } \cdot C+{ C }^{ 2 }+\left\{ \frac { { q }^{ 2 } }{ { k }^{ 2 } } +n\cdot \frac { q }{ k } \cdot C+{ C }^{ 2 } \right\} }{ n } \\ \\ =\frac { \frac { { p }^{ 2 } }{ { k }^{ 2 } } +\frac { { q }^{ 2 } }{ { k }^{ 2 } } +n\cdot \frac { 1 }{ k } \cdot C\left( p+q \right) +n{ C }^{ 2 } }{ n } \\ \\ =\frac { \frac { 1 }{ { k }^{ 2 } } \left( { p }^{ 2 }+{ q }^{ 2 } \right) +n\cdot \frac { 1 }{ k } \cdot C\left( p+q \right) +n{ C }^{ 2 } }{ n } \\ \\ =\frac { \frac { 1 }{ { k }^{ 2 } } \left( { p }^{ 2 }+{ q }^{ 2 } \right) }{ n } +\frac { n\cdot \frac { 1 }{ k } \cdot C\left( p+q \right) }{ n } +\frac { n{ C }^{ 2 } }{ n } \\ \\ =\frac { 1 }{ { k }^{ 2 } } \cdot \frac { \Sigma { x }^{ 2 } }{ n } +\frac { 1 }{ k } \cdot C\cdot \Sigma x+{ C }^{ 2 }$

Therefore, you’d have to say that: ${ \sigma }_{ y }=\sqrt { \frac { \Sigma { y }^{ 2 } }{ n } -{ \left( \frac { \Sigma y }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { \frac { 1 }{ { k }^{ 2 } } \cdot \frac { \Sigma { x }^{ 2 } }{ n } +\frac { 1 }{ k } \cdot C\cdot \Sigma x+{ C }^{ 2 }-{ \left( \frac { 1 }{ k } \cdot \frac { \Sigma x }{ n } +C \right) }^{ 2 } } \\ \\ =\sqrt { \frac { 1 }{ { k }^{ 2 } } \cdot \frac { \Sigma { x }^{ 2 } }{ n } +\frac { 1 }{ k } \cdot C\cdot \Sigma x+{ C }^{ 2 }-\left\{ \frac { 1 }{ { k }^{ 2 } } \cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 }+n\cdot \frac { 1 }{ k } \cdot \frac { \Sigma x }{ n } \cdot C+{ C }^{ 2 } \right\} } \\ \\ =\sqrt { \frac { 1 }{ { k }^{ 2 } } \cdot \frac { \Sigma { x }^{ 2 } }{ n } +\frac { 1 }{ k } \cdot C\cdot \Sigma x+{ C }^{ 2 }-\frac { 1 }{ { k }^{ 2 } } \cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 }-\frac { 1 }{ k } \cdot C\cdot \Sigma x-{ C }^{ 2 } } \\ \\ =\sqrt { \frac { 1 }{ { k }^{ 2 } } \cdot \frac { \Sigma { x }^{ 2 } }{ n } -\frac { 1 }{ { k }^{ 2 } } \cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { \frac { 1 }{ { k }^{ 2 } } \left\{ \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 } \right\} } \\ \\ =\sqrt { \frac { 1 }{ { k }^{ 2 } } } \cdot \sqrt { \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =\frac { 1 }{ k } \cdot { \sigma }_{ x }$

# Coded Data Proofs: Mean & Standard Deviation, y=kx+C

Coded Data Proofs (3):

Say y=kx+C and also that:

x={p, q} and y={kp+C, kq+C}

This would mean that: $\frac { \Sigma y }{ n } =\frac { kp+C+\left\{ kq+C \right\} }{ n } \\ \\ =\frac { kp+kq+nC }{ n } \\ \\ =\frac { k\left( p+q \right) +nC }{ n } \\ \\ =\frac { k\left( p+q \right) }{ n } +\frac { nC }{ n } \\ \\ =k\cdot \frac { \Sigma x }{ n } +C$

And if the above is true: $\frac { \Sigma { y }^{ 2 } }{ n } =\frac { { \left( kp+C \right) }^{ 2 }+{ \left( kq+C \right) }^{ 2 } }{ n } \\ \\ =\frac { \left( kp+C \right) \left( kp+C \right) +\left( kq+C \right) \left( kq+C \right) }{ n } \\ \\ =\frac { { k }^{ 2 }{ p }^{ 2 }+nkpC+{ C }^{ 2 }+\left\{ { k }^{ 2 }{ q }^{ 2 }+nkqC+{ C }^{ 2 } \right\} }{ n } \\ \\ =\frac { { k }^{ 2 }{ p }^{ 2 }+{ k }^{ 2 }{ q }^{ 2 }+nkC\left( p+q \right) +n{ C }^{ 2 } }{ n } \\ \\ =\frac { { k }^{ 2 }\left( { p }^{ 2 }+{ q }^{ 2 } \right) +nkC\left( p+q \right) +n{ C }^{ 2 } }{ n } \\ \\ =\frac { { k }^{ 2 }\left( { p }^{ 2 }+{ q }^{ 2 } \right) }{ n } +\frac { nkC\left( p+q \right) }{ n } +\frac { n{ C }^{ 2 } }{ n } \\ \\ ={ k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +kC\cdot \Sigma x+{ C }^{ 2 }\\ \\ ={ k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right)$

Therefore: ${ \sigma }_{ y }=\sqrt { \frac { \Sigma { y }^{ 2 } }{ n } -{ \left( \frac { \Sigma y }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -{ \left( k\cdot \frac { \Sigma x }{ n } +C \right) }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -\left( k\cdot \frac { \Sigma x }{ n } +C \right) \left( k\cdot \frac { \Sigma x }{ n } +C \right) } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -\left\{ { k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 }+nkC\cdot \frac { \Sigma x }{ n } +{ C }^{ 2 } \right\} } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -{ k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 }-nkC\cdot \frac { \Sigma x }{ n } -{ C }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -{ k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 }-kC\cdot \Sigma x-{ C }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -{ k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 }-C\left( k\cdot \Sigma x+C \right) } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } -{ k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\left\{ \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 } \right\} } \\ \\ =\sqrt { { k }^{ 2 } } \cdot \sqrt { \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =k\cdot { \sigma }_{ x }$

# How to add up all the even numbers from 0 onwards quickly

In this post, I’ll be demonstrating how you can add up all the even numbers from 0 onwards.

Adding up all the even numbers from 0 to 2: In this diagram, we are going to say that n=2. The height of the rectangle is (n+2) and its length is n/2. This means that the area shaded in red, which is in fact equal to all the even numbers from 0 to 2 added up, is: $\left\{ \left( n+2 \right) \cdot \frac { n }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { n\left( n+2 \right) }{ 4 }$

Adding up all the even numbers from 0 to 4: In this diagram, we are going to say that n=4. The height of the rectangle is (n+2) and its length is n/2. This means that the area shaded in red, which is in fact equal to all the even numbers from 0 to 4 added up, is: $\left\{ \left( n+2 \right) \cdot \frac { n }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { n\left( n+2 \right) }{ 4 }$

Adding up all the even numbers from 0 to 6: In this diagram, we are going to say that n=6. The height of the rectangle is (n+2) and its length is n/2. This means that the area shaded in red, which is in fact equal to all the even numbers from 0 to 6 added up, is: $\left\{ \left( n+2 \right) \cdot \frac { n }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { n\left( n+2 \right) }{ 4 }$

Adding up all the even numbers from 0 to 8: In this diagram, we are going to say that n=8. The height of the rectangle is (n+2) and its length is n/2. This means that the area shaded in red, which is in fact equal to all the even numbers from 0 to 8 added up, is: $\left\{ \left( n+2 \right) \cdot \frac { n }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { n\left( n+2 \right) }{ 4 }$

What we’ve discovered:

We’ve discovered that a simple formula can be used to add up all the even numbers from 0 to “n”, whereby “n” is an even number. This formula is: $\left\{ \left( n+2 \right) \cdot \frac { n }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { n\left( n+2 \right) }{ 4 }$

Alternative method:

There is also an alternative formula you can use to add up even numbers, from 0 onwards. That is: # How to add up odd numbers from 0 upwards

In this post, I’ll be demonstrating how to add up all the odd numbers from 0 to any specific odd number. To create a robust demonstration, I’ll be taking the footsteps below:

• I’ll first be showing you how to add up all the odd numbers from 0 to 1, using a diagram and formula.
• I’ll then be showing you how to add up all the odd numbers from 0 to 3, using a diagram and formula.
• I’ll also be showing you how to add up all the odd numbers from 0 to 5, using a diagram and also the same formula which was used to count up all the odd numbers from 0 to 1 and 0 to 3.
• And finally, I’ll be using similar diagrams and formulas used to count odd numbers from 0 to 1, 0 to 3 and 0 to 5 to count odd numbers from 0 to 7 and 0 to 9.

What you will find, after I complete the tasks above – is that a pattern emerges. You will notice that the formula I use to count odd numbers from 0 to n (n which is an odd number) is very robust and will allow you to count all the odd numbers from 0 to n very easily.

COUNTING ALL THE ODD NUMBERS FROM 0 to 1: If you count all the odd numbers from 0 to 1, what you will get is obviously 1. Furthermore, what you will also get as a formula (if n=1, H=Height and L=Length) is: $\left\{ H\cdot L \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\left\{ \left( n+1 \right) \cdot \frac { \left( n+1 \right) }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { { \left( n+1 \right) }^{ 2 } }{ 4 }$

*If you plug the value 1 into n, you will get 1. 1 is the value of all the odd numbers added up from 0 to 1.

COUNTING ALL THE ODD NUMBERS FROM 0 to 3: If you count all the odd numbers from 0 to 3, what you will get is 4. Furthermore, what you will also get as a formula (if n=3, H=Height and L=Length) is: $\left\{ H\cdot L \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\left\{ \left( n+1 \right) \cdot \frac { \left( n+1 \right) }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { { \left( n+1 \right) }^{ 2 } }{ 4 }$

*If you plug the value 3 into n, you will get 4. 4 is the value of all the odd numbers added up from 0 to 3.

COUNTING ALL THE ODD NUMBERS FROM 0 to 5: If you count all the odd numbers from 0 to 5, what you will get is 9. Furthermore, what you will also get as a formula (if n=5, H=Height and L=Length) is: $\left\{ H\cdot L \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\left\{ \left( n+1 \right) \cdot \frac { \left( n+1 \right) }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { { \left( n+1 \right) }^{ 2 } }{ 4 }$

*If you plug the value 5 into n, you will get 9. 9 is the value of all the odd numbers added up from 0 to 5.

COUNTING ALL THE ODD NUMBERS FROM 0 to 7: If you count all the odd numbers from 0 to 7, what you will get is 16. Furthermore, what you will also get as a formula (if n=7, H=Height and L=Length) is: $\left\{ H\cdot L \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\left\{ \left( n+1 \right) \cdot \frac { \left( n+1 \right) }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { { \left( n+1 \right) }^{ 2 } }{ 4 }$

*If you plug the value 7 into n, you will get 16. 16 is the value of all the odd numbers added up from 0 to 7.

COUNTING ALL THE ODD NUMBERS FROM 0 to 9: If you count all the odd numbers from 0 to 9, what you will get is 25. Furthermore, what you will also get as a formula (if n=9, H=Height and L=Length) is: $\left\{ H\cdot L \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\left\{ \left( n+1 \right) \cdot \frac { \left( n+1 \right) }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { { \left( n+1 \right) }^{ 2 } }{ 4 }$

*If you plug the value 9 into n, you will get 25. 25 is the value of all the odd numbers added up from 0 to 9.

THE FORMULA WHICH CAN BE USED TO ADD UP ALL THE ODD NUMBERS FROM 0 TO n, WHEREBY n IS AN ODD NUMBER:

If you look at each and every diagram and formula above, what you will notice is that the formula $Formula=\frac { { \left( n+1 \right) }^{ 2 } }{ 4 }$

will allow you to add up all the odd numbers from 0 to n, whereby n is an odd number. The diagrams above have demonstrated why this formula is robust and completely logical. If you need to add up all the odd numbers from 0 to n (n is an odd number), the formula above is one you can trust.

ALTERNATIVE METHOD:

Using the table below, we can come up with an alternative method of calculating every odd number from 0 to n (n is an odd number):

n: Sum Total Total (Exponential form)
1 1 1 1^2
3 1+3 4 2^2
5 1+3+5 9 3^2
7 1+3+5+7 16 4^2
9 1+3+5+7+9 25 5^2

It turns out that: *Note that 2x+1 can be used to denote an odd number.