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Question-199338




Question Number 199338 by sonukgindia last updated on 01/Nov/23
Answered by aleks041103 last updated on 01/Nov/23
5^2^M  −1=(5^2^(M−1)  )^2 −1^2 =(5^2^(M−1)  +1)(5^2^(M−1)  −1)  ((Π_(k=1) ^(M−1) (5^2^k  +1))/(5^2^M  −1))=(((5^2^(M−1)  +1)Π_(k=1) ^(M−2) (5^2^k  +1))/((5^2^(M−1)  +1)(5^2^(M−1)  −1)))=((Π_(k=1) ^(M−2) (5^2^k  +1))/(5^2^(M−1)  −1))  by induction  ⇒P=((5^2^1  +1)/(5^2^2  −1))=((5^2 +1)/((5^2 )^2 −1^2 ))=(1/(5^2 −1))  ⇒P = (1/(24))
$$\mathrm{5}^{\mathrm{2}^{{M}} } −\mathrm{1}=\left(\mathrm{5}^{\mathrm{2}^{{M}−\mathrm{1}} } \right)^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} =\left(\mathrm{5}^{\mathrm{2}^{{M}−\mathrm{1}} } +\mathrm{1}\right)\left(\mathrm{5}^{\mathrm{2}^{{M}−\mathrm{1}} } −\mathrm{1}\right) \\ $$$$\frac{\underset{{k}=\mathrm{1}} {\overset{{M}−\mathrm{1}} {\prod}}\left(\mathrm{5}^{\mathrm{2}^{{k}} } +\mathrm{1}\right)}{\mathrm{5}^{\mathrm{2}^{{M}} } −\mathrm{1}}=\frac{\left(\mathrm{5}^{\mathrm{2}^{{M}−\mathrm{1}} } +\mathrm{1}\right)\underset{{k}=\mathrm{1}} {\overset{{M}−\mathrm{2}} {\prod}}\left(\mathrm{5}^{\mathrm{2}^{{k}} } +\mathrm{1}\right)}{\left(\mathrm{5}^{\mathrm{2}^{{M}−\mathrm{1}} } +\mathrm{1}\right)\left(\mathrm{5}^{\mathrm{2}^{{M}−\mathrm{1}} } −\mathrm{1}\right)}=\frac{\underset{{k}=\mathrm{1}} {\overset{{M}−\mathrm{2}} {\prod}}\left(\mathrm{5}^{\mathrm{2}^{{k}} } +\mathrm{1}\right)}{\mathrm{5}^{\mathrm{2}^{{M}−\mathrm{1}} } −\mathrm{1}} \\ $$$${by}\:{induction} \\ $$$$\Rightarrow{P}=\frac{\mathrm{5}^{\mathrm{2}^{\mathrm{1}} } +\mathrm{1}}{\mathrm{5}^{\mathrm{2}^{\mathrm{2}} } −\mathrm{1}}=\frac{\mathrm{5}^{\mathrm{2}} +\mathrm{1}}{\left(\mathrm{5}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Rightarrow{P}\:=\:\frac{\mathrm{1}}{\mathrm{24}} \\ $$
Commented by sonukgindia last updated on 01/Nov/23
correct
$${correct} \\ $$
Answered by MathematicalUser2357 last updated on 04/Nov/23
B I G
$$\boldsymbol{\mathrm{B}}\:\boldsymbol{\mathrm{I}}\:\boldsymbol{\mathrm{G}} \\ $$

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