Question

Prove that the difference between two consecutive square numbers is always odd

Answer 1

Answer:

$(a+1)^{2} -a^{2} = \\a^{2}+1+2a-a^{2} =\\ 2a+1$

It's always odd

Step-by-step explanation:

Answer 2

✽ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ✽

➷ The first square number can be represented by $n^{2}$

The second square number can be represented by $(n+1)^{2}$

Subtract them:

$(n+1)^{2} - n^{2}$

Expand it first:

$n^{2} + 2n + 1 - n^{2}$

Simplify:

$2n+1$

Anything multiplied by 2 is even. Adding 1 to this value would make it odd. This proves it.

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡