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Chapter 6: Problem 488

For all positive integral values of \(n, 3^{3 n}-2 n+1\) is divisible by (a) 2 (b) 4 (c) 8 (d) 12

Short Answer

Expert verified
The expression \(3^{3n} - 2n + 1\) is divisible by 2 for all positive integral values of \(n\). So the correct answer is (a) 2.

Step by step solution

01

Rewrite the expression

Let's rewrite the given expression as follows: \[3^{3n} - 2n + 1 = (3^n)^3 - 2n + 1\]
02

Observe the pattern of the expression for different values of n

Let's find the value of the expression for different values of \(n\) and see if we can observe a pattern. - For \(n = 1\): \[(3^1)^3 - 2(1) + 1 = 3^3 - 2 + 1 = 27 - 1 = 26\] - For \(n = 2\): \[(3^2)^3 - 2(2) + 1 = 3^6 - 4 + 1 = 729 - 3 = 726\] - For \(n = 3\): \[(3^3)^3 - 2(3) + 1 = 3^9 - 6 + 1 = 19683 - 5 = 19678\] We can observe that each value of the expression ends in an even number, so the expression is indeed divisible by 2 for any positive integer value of \(n\). Now let's check for other divisors.
03

Test whether the expression is divisible by 4 or 8

If the expression is divisible by 4 or 8, then it must also be divisible by 2. To check if the expression is divisible by 4 or 8, let's find the remainder when the expression is divided by 4 or 8. For \(n = 1\), the expression's value was 26. When 26 is divided by 4, the remainder is 2, and when 26 is divided by 8, the remainder is 2 as well. This means that the expression is NOT divisible by 4 or 8 when \(n = 1\). Therefore, the only common divisor among the options that can evenly divide the expression for all positive integer values of \(n\) is 2.
04

Conclusion

The expression \(3^{3n} - 2n + 1\) is divisible by 2 for all positive integral values of \(n\). So the correct answer is (a) 2.

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Most popular questions from this chapter

\(\forall n \in N,\left(3+5^{(1 / 2)}\right)^{n}+\left(3-5^{(1 / 2)}\right)^{\mathrm{n}}\) is (a) Even natural number (b) Odd natural number (c) Any natural number (d) Rational number

By principle of mathematical induction, \(\forall \mathrm{n} \subset \mathrm{N}, \cos \theta \cos 2 \theta \cos 4 \theta \ldots \cos \left[\left(2^{\mathrm{n}-1}\right) \theta\right]=\) (a) \(\left[\left(\sin 2^{n} \theta\right) /\left(2^{n} \sin \theta\right)\right]\) (b) \(\left[\left(\cos 2^{n} \theta\right) /\left(2^{n} \sin \theta\right)\right]\) (c) \(\left[\left(\sin 2^{\mathrm{n}} \theta\right) /\left(2^{\mathrm{n}-1} \sin \theta\right)\right]\) (d) None of these

If \(\mathrm{x}^{2 \mathrm{n}-1}+\mathrm{y}^{2 \mathrm{n}-1}\) is divisible by \(\mathrm{x}+\mathrm{y}\), then \(\mathrm{n}\) is (a) Positive integer (b) Only for an even positive integer (c) An odd positive integer (d) \(\forall \mathrm{n}, \mathrm{n} \geq 2\)

If matrix \(\mathrm{A}=\left|\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right|\) and \(\mathrm{I}=\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|\) then which one of the following holds for all \(n \in N\). (use principle of mathematical Induction) (a) \(\mathrm{A}^{\mathrm{n}}=\mathrm{n} \cdot \mathrm{A}-(\mathrm{n}-1) \mathrm{I}\) (b) \(\mathrm{A}^{\mathrm{n}}=2^{\mathrm{n}-1} \cdot \mathrm{A}+(\mathrm{n}-1) \mathrm{I}\) (a) \(\mathrm{A}^{\mathrm{n}}=\mathrm{n} \cdot \mathrm{A}+(\mathrm{n}-1) \mathrm{I}\) (b) \(A^{n}=2^{n-1} \cdot A-(n-1) I\)

For each \(n \in N, 3^{2 n}-1\) is divisible by (a) 8 (b) 16 (c) 32 (d) None of these
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