For what value of $n$ are $2^{n} -$ 1 and $2^{n}$ + 1 Prime ?

7

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2

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1

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Solution

The correct option is C $2$
We check by substituting each of the given values
On putting n = 2 we get
$2^{2} - 1 = 3$ and $2^{2} + 1 = 5$ which are both prime numbers

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Similar questions

The value of $2^{n} {1 \times 3 \times 5..... (2 n - 3) \times (2 n - 1)}$ is $\frac{(2 n)!}{n!}$

1^{2} + 3^{2} + 5^{2} \dots + (2 n - 1)^{2} = \frac{n}{x} (2 n - 1) (2 n + 1)

n \in N

x

^{2 n} C_{1} +^{2 n} C_{3} +^{2 n} C_{5} + . . . . . +^{2 n} C_{n - 1} = 2^{2 n - 1}

^{2 n} C_{1} +^{2 n} C_{3} +^{2 n} C_{5} + . . . . . +^{2 n} C_{n - 1} = 2^{2 n - 2}

n, (n \in N)

2 n + 1

\frac{1^{2}}{(1) (3)} + \frac{2^{2}}{(3) (5)} + . . . . . + \frac{n^{2}}{(2 n - 1) (2 n + 1)} = \frac{n (n + 1)}{2 (2 n + 1)}

\frac{1}{(1) (3)} + \frac{1}{(3) (5)} + . . . . . + \frac{1}{(2 n - 1) (2 n + 1)} = \frac{1}{2 n + 1}

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