2-122-124-128-1-28-1-4 n -1- then the value of n isA- 4B- 3C- 2D- 1

The correct option is A 4We know that, a^2 b^2=(a+b)(a b) Given expression is (2+1)(2^2+1)(2^4+1)(2^8+1)(2^8 1) nbsp;= (2+1)(2^2+1)(2^4+1)(2^8+1)(2 1)(2+1)( ...

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Byju's Answer

Standard XII

Mathematics

Sum of Coefficients of All Terms

2+122+124+128...

Question

If $\frac{(2 + 1) (2^{2} + 1) (2^{4} + 1) (2^{8} + 1)}{(2^{8} - 1)} = 4^{n} + 1,$ then the value of $n$ is

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Solution

The correct option is A $4$
We know that,
$a^{2} - b^{2} = (a + b) (a - b)$
Given expression is
$\frac{(2 + 1) (2^{2} + 1) (2^{4} + 1) (2^{8} + 1)}{(2^{8} - 1)} = \frac{(2 + 1) (2^{2} + 1) (2^{4} + 1) (2^{8} + 1)}{(2 - 1) (2 + 1) (2^{2} + 1) (2^{4} + 1)} = 2^{8} + 1 = 4^{4} + 1 ∴ n = 4$

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If (2+1)(22+1)(24+1)(28+1)(28−1)=4n+1, then the value of n is

The correct option is A 4We know that, a2−b2=(a+b)(a−b) Given expression is (2+1)(22+1)(24+1)(28+1)(28−1)=(2+1)(22+1)(24+1)(28+1)(2−1)(2+1)(22+1)(24+1)=28+1=44+1∴n=4

If $\frac{(2 + 1) (2^{2} + 1) (2^{4} + 1) (2^{8} + 1)}{(2^{8} - 1)} = 4^{n} + 1,$ then the value of $n$ is

The correct option is A $4$
We know that,
$a^{2} - b^{2} = (a + b) (a - b)$
Given expression is
$\frac{(2 + 1) (2^{2} + 1) (2^{4} + 1) (2^{8} + 1)}{(2^{8} - 1)} = \frac{(2 + 1) (2^{2} + 1) (2^{4} + 1) (2^{8} + 1)}{(2 - 1) (2 + 1) (2^{2} + 1) (2^{4} + 1)} = 2^{8} + 1 = 4^{4} + 1 ∴ n = 4$