Mathematics: Algebrically prove that the difference of the squares of any two consecutive integers is an odd integer

Algebrically prove that the difference of the

Ответ:

ccdarling7185

04.08.2019

Let $n$ be the first integer. the next integer must then be $n+1$ .the difference of their squares is an odd integer, so $(n+1)^2 - n^2$ $(n+1)^2$ is expandable. $\begin{aligned} (n+1)^2 & = (n+1)(n+1) \\ & =n(n+1) + 1(n+1) \\ & = n^2 + n + n + 1 \\ & = n^2 + 2n + 1 \end{aligned}$ therefore $\begin{aligned} (n+1)^2 - n^2 & = (n^2 + 2n + 1) - n^2 \\ & = n^2- n^2 + 2n + 1 \\ & = 0 + 2n + 1\\ & = 2n + 1 \end{aligned}$ all even integers are divisible by 2. $2n$ forces $n$ to always be an even integer. it forces 2 to be a factor of $n$ .if you add 1 to any even integer, you get an odd integer.so 2n makes always an even integers ; adding 1 therefore makes it always odd.

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Ответ:

romeroalexis817

23.07.2021

Mathematics

Step-by-step explanation:

y = a * x(x - 3)(x + 1)^2

2,12 is used to find the value of a

y = a (2)(2 - 3)(2 + 1)^2

12 = a (2)(-1)(3)^2

12 = a (2) (-1) (9)

12 = - 18*a

12/-18 = a

a = - 2/3

y = (-2/3)*x*(x - 3)(x + 1)^2

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Algebrically prove that the difference of the squares of any two consecutive integers is an odd integer

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