( 5^2 + 1) ( 5^4+1) (5^8+1)(5^16 +1)(5^32 + 1)^2
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b: A=1/3+1/9+...+1/3^10
=>3A=1+1/3+...+1/3^9
=>A*2=1-1/3^10=(3^10-1)/3^10
=>A=(3^10-1)/(2*3^10)
c: C=3/2+3/8+3/32+3/128+3/512
=>4C=6+3/2+...+3/128
=>3C=6-3/512
=>C=1023/512
d: A=1/2+...+1/256
=>2A=1+1/2+...+1/128
=>A=1-1/256=255/256
\(A=\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
...
\(4A=5^{64}-1\)
\(\Rightarrow A=\frac{5^{64}-1}{4}>B=\frac{5^{64}-1}{5}\)
\(12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(=\frac{1}{2}\left(5^{32}-1\right)\left(5^{32}+1\right)\)
\(=\frac{1}{2}\left(5^{64}-1\right)\)
c: \(=\dfrac{3}{2}\cdot1-1-20=\dfrac{3}{2}-21=\dfrac{-39}{2}\)