Find the value of this

[haughty yarrow]

$\left(\prod_{n=1}^{256}(2^n +1)+1\right)^{\frac{1}{256}}$

[eager venture]

Dawg

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[haughty yarrow]

what i thought of was converting $\prod_{n=1}^{256}(2^n +1)+1$ into $2^x+2$ where x is some integer

[eager venture]

Dawg

[haughty yarrow]

now i was thinking of solving as a limit from here so

$\lim_{x \to \infty} (2^x+2)^\frac{1}{x}$

[eager venture]

Dawg

[haughty yarrow]

solving this limit gives us 2

so im thinking the answer of the original question

should be near about 2?

[zenith spindle]

wat 💀

how is this even convertible

[haughty yarrow]

well it probably isnt

what my thought process was since its multiplication so the term would grow exponentially hence the 2^x. And the mulitplication thing clearly gives an odd number so 2^x+1

and there is another 1 after that

so 2^x +2

very oversimplified way of thinking of this question ig

and as for the infinity

2^x +1 is a big number

and so is the 1/256(kinda)

so making the x tend to infinity will replicate what we have in the question

[tawny burrow]

no

to both the logics

theyre both incorrect !

[tawny burrow]

thanks to coffey

trust

find a closed form for $\prod_{r=1}^{n}(2^r+1)$

[eager venture]

waffle

[tawny burrow]

maybe calculate the first few values

and induct

[tawny burrow]

@haughty yarrow

[haughty yarrow]

+close