Hi Everyone .

I had been working out on numbers for quite some time and I had got some interesting relations among numbers which I will write below .

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*Relation 1 :*

For any three consecutive integers \(a, b, c\)

**\(\color{blue}\boxed{b^2 = (a \times c) + 1}\)**

Example : let us take the numbers 24, 25, 26

\(25^2 = 24 \times 26 + 1 \Rightarrow 625 = 624 + 1 \Rightarrow 625 = 625\)

*Relation 2 :*

For any three consecutive integers \(a, b, c\)

*\(\color{red}\boxed{(a \times b) + 2c + a = c^2}\)*

Example : let us take 3 numbers 3, 4, 5

\((3 \times 4) + 2 \times 5 + 3 = 5^2 \Rightarrow 12 + 10 + 3 = 25 \Rightarrow 25 = 25\)

*Relation 3 :*

For any two consecutive integers \(a, b\)

*\(\color{yellow}\boxed{(a \times b) - a = a^2}\)*

Example : let us take two numbers 6, 7

\((6 \times 7) - 6 = 42 - 6 = 36 = 6^2\)

*Relation 4 :*

For any three consecutive numbers \(a, b, c\)

\(\color{orange}\boxed{a^2 - b^2 + 2c = 3}\) (OR) \(\color{orange}\boxed{(a + b)(a - b) + 2c = 3}\)

Example : let us take three numbers 7, 8, 9

\((7 + 8)(7 - 8) + 2(9) = -15 + 18 = 3\)

*Relation 5 :*

For any two consecutive even or odd numbers \(a, b\)

\(\color{green}\boxed{a^2 + 2a + 2b = b^2}\) (OR) \(\color{green}\boxed{a^2 + 4a + 4 = b^2}\)

Example : let us take two consecutive odd numbers 13, 15

\(13^2 + 2(13) + 2(15) = 169 + 26 + 30 = 225\)

*Relation 6 :*

For any two consecutive numbers \(a, b\) such that \(a > b\)

\(\color{brown}\boxed{a^2 - b^2 = a + b}\) (OR) \(\color{brown}\boxed{b^2 = a^2 - (a + b)}\) (OR) \(\color{violet}\boxed{a^2 = b^2 + a + b}\)

Example : let us take two numbers 31, 30

\(31^2 - 30^2 = 961 - 900 = 61 = 31 + 30\)

*Relation 7 :*

If there are three whole numbers \(a, b, c\) such that \(\color{blue}a \times b = c\) , then

\[\begin{align} a^2 \times b^2 = c^2\\ a^3 \times b^3 = c^3\\ a^4 \times b^4 = c^4\\ a^5 \times b^5 = c^5\\ \end{align}\]

So , if \(\color{violet}a \times b = c\) then \(\color{green}\boxed{a^n \times b^n = c^n}\)

Example : let us take the numbers 2, 3, 6 because \(2 \times 3 = 6\)

So , now \[\begin{align} 2^2 \times 3^2 = 6^2 = 36\\ 2^3 \times 3^3 = 6^3 = 216\\ 2^4 \times 3^4 = 6^4 = 1296\\ 2^5 \times 3^5 = 6^5 = 7776\\ \end{align}\]

*Relation 8 :*

Sum of any \(n\) consecutive integers is given by :

\(\color{red}\boxed{Sum = Median \times n}\)

Example : find the sum of 11, 12, 13, 14, 15

n = 5 (odd)

Median = \(\frac{5 + 1}{2} = \frac{6}{2} = 3^{th}\) *term = 13*

\(Sum = median \times n = 13 \times 5 = 65\)

That's all for now . Hope these relations will help you . If any mistakes are there please inform me .

If you too have any kind of relations with you please put it in form of a comment below and I will write them in this note in your name .

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