If $2^{n} - 1$ is divisible by $7$ , then $n$ must be of the form

3 r - 1

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3 r

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7 r + 1

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7 r

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Solution

The correct option is B $3 r$
$(2^{n}) - 1$
$= (2^{3})^{\frac{n}{3}} - 1$
$= (8)^{\frac{n}{3}} - 1$
$= (1 + 7)^{\frac{n}{3}} - 1$
$= 1 + 7 m - 1$
$= 7 m$ where m is positive integer.
Hence for $2^{n} - 1$ to be divisible by 7, n must has to be a multiple of 3.

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n \sum r = 0 (- 1)^{r}^{n} C_{r} [\frac{1}{2^{r}} + \frac{3^{r}}{2^{2 r}} + \frac{7^{r}}{2^{3 r}} + \dots \infty]

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n \sum r = 0 (- 1)^{r}^{n} C_{r} [\frac{1}{2^{r}} + \frac{3^{r}}{2^{2 r}} + \frac{7^{r}}{2^{3 r}} + \dots \infty]

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n \sum r = 0 {(- 1)}^{r}^{n} C_{r} [\frac{1}{2^{r}} + \frac{3^{r}}{2^{2 r}} + \frac{7^{r}}{2^{3 r}} + \frac{15^{r}}{2^{4 r}} + \dots u p t o p t e r m s]

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If 2n−1 is divisible by 7, then n must be of the form

The correct option is B 3r(2n)−1=(23)n3−1=(8)n3−1=(1+7)n3−1=1+7m−1=7m where m is positive integer.Hence for 2n−1 to be divisible by 7, n must has to be a multiple of 3.

If $2^{n} - 1$ is divisible by $7$ , then $n$ must be of the form

The correct option is B $3 r$
$(2^{n}) - 1$
$= (2^{3})^{\frac{n}{3}} - 1$
$= (8)^{\frac{n}{3}} - 1$
$= (1 + 7)^{\frac{n}{3}} - 1$
$= 1 + 7 m - 1$
$= 7 m$ where m is positive integer.
Hence for $2^{n} - 1$ to be divisible by 7, n must has to be a multiple of 3.