
If the expression ${2^{3n}} - 7n - 1$ is divisible by 49. Hence show that ${2^{3n + 3}} - 7n - 8$ is divisible by 49, $n \in N$.
Answer
629.7k+ views
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

Chemical formula of Bleaching powder is A Ca2OCl2 B class 11 chemistry CBSE

Name the part of the brain responsible for the precision class 11 biology CBSE

The growth of tendril in pea plants is due to AEffect class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

