a,1+1=
b,2+2=
c,3+3=
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a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)
a,1+1=2
b,2+2=4
c,3+3=6
hok tok
a) 1+1=2
b)2+2=4
c)3+3=6