Question

For all positive integers n, \[{3^{2n}} - 2n + 1\] is divisible by
A. 2
B. 4
C. 8
D. 12

Answer 1

Hint:We will have to start the question by hit trial. That is we have to put values for different n and check for divisibility and further can be proved by principle of mathematical induction.

Complete step-by-step answer:
We have to check whether \[{3^{2n}} - 2n + 1\] is divisible by any of the options for all values of n which belong to natural numbers.
We will start by putting 1 in place of n and check the divisibility –
=\[{3^{2n}} - 2n + 1\]
=\[{3^{2(1)}} - 2(1) + 1\]
=\[9 - 2 + 1\]
=\[8\]
Therefore, it is divisible by 2, 4 and 8 from the options.
Now, we will put 2 in place of n, we get –
=\[{3^{2n}} - 2n + 1\]
=\[{3^{2(2)}} - 2(2) + 1\]
=\[81 - 4 + 1\]
=\[78\]
This is divisible by only 2.
Therefore, 2 is the correct answer.

So, the correct answer is “Option A”.

Note:This question can be solved by an alternative method by principle of mathematical induction.
Consider $f(n)$ = \[{3^{2n}} - 2n + 1\]
Let us check if it is divisible by 2 for that it must be true for n=1,
=\[{3^{2n}} - 2n + 1\]
=\[{3^{2(1)}} - 2(1) + 1\]
=\[9 - 2 + 1\]
=\[8\]
Therefore, it is divisible by 2.
 Let us assume that \[{3^{2n}} - 2n + 1\] is divisible by 2 for any value of n=m.
=\[{3^{2n}} - 2n + 1\]
For n=m we have,
$F(m)$ =\[{3^{2(m)}} - 2(m) + 1\] = $2p……….. (1)$ (Since, it is assumed to be divisible by 2)
Now, we check for n=m+1,
$F(m+1)$ =\[{3^{2(m + 1)}} - 2(m + 1) + 1\]
$F(m+1)$ =\[{9.3^{2}} - 2(m) - 1\]
Rearranging and adding 1 and subtracting 1 we get,
=\[{3^{2n}} - 2(n) - 1 + ( - 1 - 1) + {8.3^{2n}}\]
=\[{3^{2n}} - 2(n) - 1 + {8.3^{2n}} - 2\]
From (1) we have,
=\[2p + 2({4.3^{2n}} - 1)\]
=\[2[p + ({4.3^{2n}} - 1)]\]
From above equation it is divisible by 2
Since, F(m + 1) is true, whenever F(m) is true.
Thus, F(1) is true and F(k + 1) is true, whenever F(k) is true.
Hence, by the principle of mathematical induction, F(n) is true for all n ∈ N.