Question

By the method of induction prove that \[\left( {{2}^{3n}}-1 \right)\] is divisible by 7.

Answer 1

Hint: To solve this question, we will assume P(n) be a statement such that \[{{2}^{3n}}-1\] is divisible by 7. After proving P(1) = true by taking n = 1, we will assume that P(k) is true. Finally, from this, P(k) is a true statement and we will try to get that P(k + 1) is also true.

Complete step-by-step answer:
We have to prove that \[{{2}^{3n}}-1\] is divisible by 7 with the help of mathematical induction. Let P(n) be a statement such that \[P\left( n \right)={{2}^{3n}}-1\] is divisible by 7. To show n = 1 is true, P(n) = P(1) and substituting n = 1 in \[{{2}^{3n}}-1.\]
\[\Rightarrow {{2}^{3\times 1}}-1\]
\[\Rightarrow {{2}^{3}}-1\]
\[\Rightarrow 8-1\]
\[\Rightarrow 7\]
And as 7 is divisible by 7. Therefore, \[{{2}^{3}}-1\] is divisible by 7. Hence P(1) is true as it is stated \[{{2}^{3}}-1\] is divisible by 7…..(i)
Let us assume P(k) is true where n = k and P(k) is true.
\[\Rightarrow {{2}^{3k}}-1\] is divisible by 7.
\[\Rightarrow P\left( k \right)={{2}^{3k}}-1=7d\]
It means it can be written as a multiple of 7 where d is any natural number.
\[\Rightarrow {{2}^{3k}}-1=7d\]
\[\Rightarrow {{2}^{3k}}=7d+1.....\left( iii \right)\]
where d = 1, 2, 3, 4…..
Now, we finally check for n = R + 1.
\[P\left( k+1 \right)={{2}^{3\left( k+1 \right)}}-1\]
For P(k + 1) to be true, we should have,
\[{{2}^{3\left( k+1 \right)}}-1=7d\]
That is we should have, \[{{2}^{3\left( k+1 \right)}}=7d+1.\]
Now, \[{{2}^{3\left( k+1 \right)}}={{2}^{3k}}{{.2}^{3}}\]
As \[{{a}^{xy}}={{a}^{x}}{{a}^{y}}\] where ‘a’ is any real number and x and y are real numbers or integers.
\[\Rightarrow P\left( k+1 \right)={{2}^{3k}}{{.2}^{3}}-1\]
Now from equation (iii), we had that,
\[{{2}^{3k}}=7d+1\]
Substituting this value of \[{{2}^{3k}}\] in the above equation, we have,
\[P\left( k+1 \right)=\left( 7d+1 \right){{.2}^{3}}-1\]
\[\Rightarrow P\left( k+1 \right)=\left( 7d+1 \right)8-1\]
\[\Rightarrow P\left( k+1 \right)=56d+8-1\]
\[\Rightarrow P\left( k+1 \right)=56d+7\]
\[\Rightarrow P\left( k+1 \right)=7\left( 8d+1 \right)\]
Hence, P(k + 1) is written as a product of 7 and (8d + 1) where d = 1, 2, 3…. Hence, we have, \[P\left( k+1 \right)={{2}^{3\left( k+1 \right)}}-1\] is divisible by 7. So, P(k + 1) is true…..(iv)
So, finally from equation (i), (ii) and (iv), we have that P(1), P(k) and P(k + 1); k > 1 is true.
Hence, by mathematical induction, we have shown that \[{{2}^{3n}}-1\] is divisible by 7.
Hence proved.

Note: The point where we have written that for P(k + 1) to be true, we should get \[{{2}^{3\left( k+1 \right)}}-1=7d\] type. Here, this ‘d’ can also be replaced by any other natural number ‘s’ because anyways we are going to get any number with a multiple of 7. We have used ‘d’ here as it was used earlier with P(k). Anyways, it can be shifted if required. Because clearly, \[{{2}^{3k}}-1\ne {{2}^{3\left( k+1 \right)}}-1.\] So, we can change the variable too.