Question

let n=(2+1)(2^2+1)(2^4+1)....(2^32+1)+1 then n=2^x find x

Answer 1

Answer.

• The value of x is 64.

Steps.

Given –

$\tt \implies n = (2 + 1)( {2}^{2} + 1)...( {2}^{32} + 1) + 1$

We can also write it as,

$\tt \implies n = \dfrac{2 - 1}{2 - 1} \times (2 + 1)( {2}^{2} + 1)...( {2}^{32} + 1) + 1$

$\tt \implies n = \dfrac{ {2}^{1} - 1}{2 - 1} \times (2 + 1)( {2}^{2} + 1)...( {2}^{32} + 1) + 1$

$\tt \implies n = ( {2}^{1} - 1)( {2}^{1} + 1)( {2}^{2} + 1)...( {2}^{32} + 1) + 1$

Using identity (a + b)(a - b) = a² - b², we get,

$\tt \implies n =( {2}^{2} - 1)( {2}^{2} + 1)...( {2}^{32} + 1) + 1$

Again, using same identity, we get,

$\tt \implies n =( {2}^{4} - 1)( {2}^{4} + 1)...( {2}^{32} + 1) + 1$

This goes on till the last term. So, we get,

$\tt \implies n = ( {2}^{32} - 1) ( {2}^{32} + 1) + 1$

$\tt \implies n = {2}^{64} - 1 + 1$

$\tt \implies n = {2}^{64}$

It's given that,

$\tt \implies n = {2}^{x}$

$\tt \implies {2}^{64} = {2}^{x}$

Comparing base, we get,

$\tt \implies x = 64$

★ So, the value of x is 64.