Subbu's square formula

I am here to talk about square of consecutive natural numbers . $formula=>(n+1)^2-n^2=2n+1$ suppose when we consider $8^2and9^2$ then the difference between them is 2(8)+1=16+1=17 let us check this by normal method $8^2=64;9^2=81$ the difference between them is 81-64=17 that's it i got the answer this is simple because square numbers always differ by odd numbers so 2n+1 is the general form of odd numbers and i compared and got the simple Subbu's square formula please like my note no one is liking my problems and note please if you like it nothing happens to you

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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Comments

This is the general form of odd numbers.This was found before itself.This is not Subbu's square formula.

Vishal S - 6 years, 4 months ago

liked it

SARAN .P.S - 5 years, 7 months ago

This is a good note i am inspired by you

subramanyam ravichandran - 6 years, 4 months ago

thank you my boy

sudoku subbu - 6 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$