3 years ago

Prove by mathematical induction. 2n<(n+2)!, for n ≥ 0, where n ∈ ℤ

Mathematics

1 answer:

Mekhanik [1.2K]3 years ago

6 0

The question is difficult for a beginner like me

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(1/4)^3z-1 =16^z+2*64^z-2 Z=___ Please help

wariber [46]

Answer:

Z = 0.198877274

Step-by-step explanation:

$(\frac{1}{4})^{3z-1} = 16^z + 2*16^{z-2}\\4^{1-3z} = 4^{2z} + 4^{\frac{1}{2}}*4^{2z-4}\\4^{1-3z} = 4^{2z} + 4^{2z-4+\frac{1}{2}}\\4^{1-3z} = 4^{2z} + 4^{2z-\frac{7}{2}}\\4^{1-3z} = 4^{2z} *(1+ 4^{-\frac{7}{2}})\\4^{1-3z} = 4^{2z} *(1+ 2^{-7})\\4^{1-3z} = 4^{2z} *(1+ \frac{1}{128} )\\4^{1-3z} = 4^{2z} *(\frac{129}{128} )\\Taking\;\; Logarithm\;\; with\;\; base\;\; 4\\Log_4(4^{1-3z}) = Log_4(4^{2z}) + Log_4(\frac{129}{128})\\1-3z = 2z + 0.005613627712 \\5z = 0.994386372\\z = 0.198877274$