You visited us 1 times! Enjoying our articles? Unlock Full Access!

Geometric Progression

If n>1, the v...

Question

If $n > 1$ , the values of the positive integer $m$ for which $n^{m} + 1$ divides $a = 1 + n + n^{2} + \dots \dots + n^{63}$ is/are

8

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

16

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

32

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

64

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $32$
$a = 1 + n + n^{2} + \dots + n^{63}$
$\Rightarrow a = \frac{1 (n^{64} - 1)}{n - 1} \Rightarrow a = \frac{(n^{32} - 1) (n^{32} + 1)}{n - 1} \Rightarrow a = \frac{(n^{16} - 1) (n^{16} + 1) (n^{32} + 1)}{n - 1}$
$∴ a = (n + 1) (n^{2} + 1) (n^{4} + 1) (n^{8} + 1) (n^{16} + 1) (n^{32} + 1)$

Suggest Corrections

Similar questions

n > 1

m

n^{m} + 1

a = 1 + n + n^{2} + \dots \dots + n^{63}

n > 1

m

n^{m} + 1

a = 1 + n + n^{2} + . . . + n^{63}

n > 1

m

n^{m} + 1

1 + n + n^{2} . . . . . n^{125}

n (>)

m

(n^{m} + 1)

1 + n + n^{2} + . . . n^{225}

Let $n > 1$ be a positive integer, then find the largest integer m such that ( $n^{m}$ + 1) divides $1 + n + n^{2}$ + ......+ $n^{127}$ .

Join BYJU'S Learning Program