Answer

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Hint: A number is said to be an odd number if it is not completely divisible by 2. The odd numbers can be represented in the form of 2r+1. Since it is given that n is an odd positive integer, we can substitute n = 2r+1 where r be any integer greater than or equal to 0.

Complete step-by-step answer:

In the question, we are asked to show that ${{n}^{2}}-1$ is divisible by 8, if n is an odd positive integer. Any odd number can be represented by 2r+1. Since it is given that this odd number is positive, we can substitute n = 2r+1 where r should be an integer and it should be greater than or equal to 0.

Substituting n = 2r+1 in ${{n}^{2}}-1$, we get,

${{\left( 2r+1 \right)}^{2}}-1...............\left( 1 \right)$

We have a formula,

${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$

Using this formula in expression $\left( 1 \right)$, we get,

$\begin{align}

& 4{{r}^{2}}+1+4r-1 \\

& \Rightarrow 4{{r}^{2}}+4r \\

& \Rightarrow 4r\left( r+1 \right) \\

\end{align}$

So, we have to show that 4r(r+1) is divisible by 8 where, r is an integer greater than or equal to 0. So, we will substitute different values of r in 4r(r+1) and then check whether the numbers are divisible by 8 or not.

Substituting r=1, we get, 4r(r+1) is equal to,

4 (1) (1+1) = 8 which is divisible by 8.

Substituting r=2, we get, 4r(r+1) is equal to,

4 (2) (2+1) = 8 (3) which is divisible by 8.

Substituting r=3, we get, 4r(r+1) is equal to,

4 (3) (3+1) = 8 (6) which is divisible by 8.

Substituting r=4, we get, 4r(r+1) is equal to,

4 (4) (4+1) = 8 (10) which is divisible by 8.

We can observe that for any value of r, 4r(r+1) = 8x where x is any integer. So, we can say that 4r(r+1) is divisible by 8.

Since we substituted ${{n}^{2}}-1$ as 4r(r+1), hence, we can say that ${{n}^{2}}-1$ is divisible by 8 where n is an odd positive integer.

Note: One can also do this question by directly substituting n as odd and positive integers instead of first substituting n = 2r+1 and then substitute the values of r. This method will take a comparatively lesser amount of time.

Complete step-by-step answer:

In the question, we are asked to show that ${{n}^{2}}-1$ is divisible by 8, if n is an odd positive integer. Any odd number can be represented by 2r+1. Since it is given that this odd number is positive, we can substitute n = 2r+1 where r should be an integer and it should be greater than or equal to 0.

Substituting n = 2r+1 in ${{n}^{2}}-1$, we get,

${{\left( 2r+1 \right)}^{2}}-1...............\left( 1 \right)$

We have a formula,

${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$

Using this formula in expression $\left( 1 \right)$, we get,

$\begin{align}

& 4{{r}^{2}}+1+4r-1 \\

& \Rightarrow 4{{r}^{2}}+4r \\

& \Rightarrow 4r\left( r+1 \right) \\

\end{align}$

So, we have to show that 4r(r+1) is divisible by 8 where, r is an integer greater than or equal to 0. So, we will substitute different values of r in 4r(r+1) and then check whether the numbers are divisible by 8 or not.

Substituting r=1, we get, 4r(r+1) is equal to,

4 (1) (1+1) = 8 which is divisible by 8.

Substituting r=2, we get, 4r(r+1) is equal to,

4 (2) (2+1) = 8 (3) which is divisible by 8.

Substituting r=3, we get, 4r(r+1) is equal to,

4 (3) (3+1) = 8 (6) which is divisible by 8.

Substituting r=4, we get, 4r(r+1) is equal to,

4 (4) (4+1) = 8 (10) which is divisible by 8.

We can observe that for any value of r, 4r(r+1) = 8x where x is any integer. So, we can say that 4r(r+1) is divisible by 8.

Since we substituted ${{n}^{2}}-1$ as 4r(r+1), hence, we can say that ${{n}^{2}}-1$ is divisible by 8 where n is an odd positive integer.

Note: One can also do this question by directly substituting n as odd and positive integers instead of first substituting n = 2r+1 and then substitute the values of r. This method will take a comparatively lesser amount of time.

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