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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))

52M−1=(52M−1)2−12=(52M−1+1)(52M−1−1)∏M−1k=1(52k+1)52M−1=(52M−1+1)∏M−2k=1(52k+1)(52M−1+1)(52M−1−1)=∏M−2k=1(52k+1)52M−1−1byinduction⇒P=521+1522−1=52+1(52)2−12=152−1⇒P=124

Commented by sonukgindia last updated on 01/Nov/23

correct

correct

Answered by MathematicalUser2357 last updated on 04/Nov/23

B I G

BIG

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