Solveeit Logo
Login

Question

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

For all positive integers n, 32n−2n+1{3^{2n}} - 2n + 1 is divisible by
A. 2
B. 4
C. 8
D. 12

Explanation

Solution

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 32n−2n+1{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 –
=32n−2n+1{3^{2n}} - 2n + 1
=32(1)−2(1)+1{3^{2(1)}} - 2(1) + 1
=9−2+19 - 2 + 1
=88
Therefore, it is divisible by 2, 4 and 8 from the options.
Now, we will put 2 in place of n, we get –
=32n−2n+1{3^{2n}} - 2n + 1
=32(2)−2(2)+1{3^{2(2)}} - 2(2) + 1
=81−4+181 - 4 + 1
=7878
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)f(n) = 32n−2n+1{3^{2n}} - 2n + 1
Let us check if it is divisible by 2 for that it must be true for n=1,
=32n−2n+1{3^{2n}} - 2n + 1
=32(1)−2(1)+1{3^{2(1)}} - 2(1) + 1
=9−2+19 - 2 + 1
=88
Therefore, it is divisible by 2.
Let us assume that 32n−2n+1{3^{2n}} - 2n + 1 is divisible by 2 for any value of n=m.
=32n−2n+1{3^{2n}} - 2n + 1
For n=m we have,
F(m)F(m) =32(m)−2(m)+1{3^{2(m)}} - 2(m) + 1 = 2p………..(1)2p……….. (1) (Since, it is assumed to be divisible by 2)
Now, we check for n=m+1,
F(m+1)F(m+1) =32(m+1)−2(m+1)+1{3^{2(m + 1)}} - 2(m + 1) + 1
F(m+1)F(m+1) =9.32−2(m)−1{9.3^{2}} - 2(m) - 1
Rearranging and adding 1 and subtracting 1 we get,
=32n−2(n)−1+(−1−1)+8.32n{3^{2n}} - 2(n) - 1 + ( - 1 - 1) + {8.3^{2n}}
=32n−2(n)−1+8.32n−2{3^{2n}} - 2(n) - 1 + {8.3^{2n}} - 2
From (1) we have,
=2p+2(4.32n−1)2p + 2({4.3^{2n}} - 1)
=2[p+(4.32n−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.