# If the expression ${2^{3n}} - 7n - 1$ is divisible by 49. Hence show that ${2^{3n + 3}} - 7n - 8$ is divisible by 49, $n \in N$.

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Hint: Try to simplify the required expression which is to be proved to be divisible by 49 in terms of 49n and the given expression ${2^{3n}} - 7n - 1$.

Complete step-by-step answer:

It is given that ${2^{3n}} - 7n - 1$ is divisible by 49 and we need to show that ${2^{3n + 3}} - 7n - 8$ is also divisible by 49. Let’s first understand the meaning of $n \in N$, this means that the divisibility by 49 must hold true for all n which belong to the set of natural numbers. Now if an expression is proved to have dependence upon 49n that it is clear that the expression will also be divisible by 49 as 49n is divisible by 49 for all $n \in N$. Thus try to simplify the required expression which is to be proved to be divisible by 49 in terms of 49n and the given expression ${2^{3n}} - 7n - 1$. If this dependency holds true that we get the answer.

It is given that ${2^{3n}} - 7n - 1$ is divisible by 49.

Let $y = {2^{3n}} - 7n - 1$ ………………. (1)

Now we have to show that ${2^{3n + 3}} - 7n - 8$ is also divisible by 49$n \in N$.

Proof:

Let $q = {2^{3n + 3}} - 7n - 8$

Now first simplify q we have,

$

\Rightarrow q = {2^{3n}}{.2^3} - 7n - 8 \\

\Rightarrow q = {8.2^{3n}} - 7n - 8 \\

\Rightarrow q = \left( {7 + 1} \right){2^{3n}} - 7n - 1 - 7 \\

\Rightarrow q = {7.2^{3n}} - 7 + {2^{3n}} - 7n - 1 \\

$

$ \Rightarrow q = 7\left( {{2^{3n}} - 1} \right) + {2^{3n}} - 7n - 1$…………… (2)

Now from equation (1)

$y = {2^{3n}} - 7n - 1$

$ \Rightarrow {2^{3n}} - 1 = y - 7n$

Now substitute these values in equation (2) we have,

$

\Rightarrow q = 7\left( {y - 7n} \right) + y \\

\Rightarrow q = 7y - 49n + y \\

$

$ \Rightarrow q = 8y - 49n$

Now as we see that 49n is divisible by 49 because n belongs to natural number (i.e. $n \in N$)

And from equation (1) it is given that y is divisible by 49.

Therefore both 8y and 49n are divisible by 49.

Therefore $\left( {8y - 49n} \right)$ is also divisible by 49.

But, $q = 8y - 49n$ and q is equal to$\left( {7\left( {{2^{3n}} - 1} \right) + {2^{3n}} - 7n - 1} \right)$.

Therefore $\left( {7\left( {{2^{3n}} - 1} \right) + {2^{3n}} - 7n - 1} \right)$ is divisible by 49.

Hence proved.

Note: Whenever we face such types of problems the key concept is to understand the dependency and the set of values on which n belongs. Sometimes the two expressions given will not have the same set of values of n, thus we need to take care that the same set of values is being dealt with. This concept will help you get the right solution.

Complete step-by-step answer:

It is given that ${2^{3n}} - 7n - 1$ is divisible by 49 and we need to show that ${2^{3n + 3}} - 7n - 8$ is also divisible by 49. Let’s first understand the meaning of $n \in N$, this means that the divisibility by 49 must hold true for all n which belong to the set of natural numbers. Now if an expression is proved to have dependence upon 49n that it is clear that the expression will also be divisible by 49 as 49n is divisible by 49 for all $n \in N$. Thus try to simplify the required expression which is to be proved to be divisible by 49 in terms of 49n and the given expression ${2^{3n}} - 7n - 1$. If this dependency holds true that we get the answer.

It is given that ${2^{3n}} - 7n - 1$ is divisible by 49.

Let $y = {2^{3n}} - 7n - 1$ ………………. (1)

Now we have to show that ${2^{3n + 3}} - 7n - 8$ is also divisible by 49$n \in N$.

Proof:

Let $q = {2^{3n + 3}} - 7n - 8$

Now first simplify q we have,

$

\Rightarrow q = {2^{3n}}{.2^3} - 7n - 8 \\

\Rightarrow q = {8.2^{3n}} - 7n - 8 \\

\Rightarrow q = \left( {7 + 1} \right){2^{3n}} - 7n - 1 - 7 \\

\Rightarrow q = {7.2^{3n}} - 7 + {2^{3n}} - 7n - 1 \\

$

$ \Rightarrow q = 7\left( {{2^{3n}} - 1} \right) + {2^{3n}} - 7n - 1$…………… (2)

Now from equation (1)

$y = {2^{3n}} - 7n - 1$

$ \Rightarrow {2^{3n}} - 1 = y - 7n$

Now substitute these values in equation (2) we have,

$

\Rightarrow q = 7\left( {y - 7n} \right) + y \\

\Rightarrow q = 7y - 49n + y \\

$

$ \Rightarrow q = 8y - 49n$

Now as we see that 49n is divisible by 49 because n belongs to natural number (i.e. $n \in N$)

And from equation (1) it is given that y is divisible by 49.

Therefore both 8y and 49n are divisible by 49.

Therefore $\left( {8y - 49n} \right)$ is also divisible by 49.

But, $q = 8y - 49n$ and q is equal to$\left( {7\left( {{2^{3n}} - 1} \right) + {2^{3n}} - 7n - 1} \right)$.

Therefore $\left( {7\left( {{2^{3n}} - 1} \right) + {2^{3n}} - 7n - 1} \right)$ is divisible by 49.

Hence proved.

Note: Whenever we face such types of problems the key concept is to understand the dependency and the set of values on which n belongs. Sometimes the two expressions given will not have the same set of values of n, thus we need to take care that the same set of values is being dealt with. This concept will help you get the right solution.

Last updated date: 20th Sep 2023

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