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Question Number 2748 by Rasheed Soomro last updated on 26/Nov/15
Show that ((3^(2n+1) +5^(2n+1) )/8) is an integer for n∈Z^+ .
Showthat32n+1+52n+18isanintegerforn∈Z+.
Answered by prakash jain last updated on 26/Nov/15
f(n)=3^(2n+1) +5^(2n+1) f(1)=27+125=152=19×8 Let us say f(n) 3^(2n+1) +5^(2n+1) is divisible by 8. f(n+1)=3^(2n+3) +5^(2n+3) f(n+1)−f(n)=3^(2n+3) −3^(2n+1) +5^(2n+3) −5^(2n+1) =3^(2n+1) (8)+5^(2n+1) (24) =8(3^(2n+1) +3∙5^(2n+1) ) ∵ ((f(n))/8)=k⇒f(n)=8k, k∈N f(n+1)=8(3^(2n+1) +3∙5^(2n+1) )+8k f(n+1)=8(3^(2n+1) +3∙5^(2n+1) +k) ∴ f(n+1) is divisible by 8 or ((f(n+1))/8) is an integer.
f(n)=32n+1+52n+1f(1)=27+125=152=19×8Letussayf(n)32n+1+52n+1isdivisibleby8.f(n+1)=32n+3+52n+3f(n+1)−f(n)=32n+3−32n+1+52n+3−52n+1=32n+1(8)+52n+1(24)=8(32n+1+3⋅52n+1)∵f(n)8=k⇒f(n)=8k,k∈Nf(n+1)=8(32n+1+3⋅52n+1)+8kf(n+1)=8(32n+1+3⋅52n+1+k)∴f(n+1)isdivisibleby8orf(n+1)8isaninteger.
Commented by RasheedAhmad last updated on 26/Nov/15
Nice!
Nice!

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