Question

BT7: Tính

\(3,C=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{16}+1\right)\)

\(4,D=15\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)

\(5,E=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{128}+1\right)+\left(5^{256}-1\right)\)

Answer 1

3: =(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)

=(5^4-1)(5^4+1)(5^8+1)(5^16+1)

=(5^8-1)(5^8+1)(5^16+1)

=(5^16-1)(5^16+1)

=5^32-1

4:

D=(4^4-1)(4^4+1)(4^8+1)*....*(4^64+1)

=(4^8-1)(4^8+1)*...*(4^64+1)

=...

=4^128-1

5: =(5^2-1)(5^2+1)(5^4+1)*...*(5^128+1)+(5^256-1)

=(5^4-1)(5^4+1)*...*(5^128+1)+5^256-1

=5^256-1+5^256-1

=2*5^256-2

Answer 2

3, \(C=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)

\(C=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)

\(C=\left(5^4-1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)

\(C=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(C=\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(C=5^{32}-1\)

4, \(D=15\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^4-1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^8-1\right)\left(4^8+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^{16}-1\right)\left(4^{16}+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^{32}-1\right)\left(4^{32}+1\right)\left(4^{64}+1\right)\)

\(D=\left(4^{64}-1\right)\left(4^{64}+1\right)\)

\(D=4^{128}-1\)

5, \(E=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{256}+1\right)\)

\(E=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{128}+1\right)\left(5^{256}+1\right)\)

\(E=\left(5^4-1\right)\left(5^4+1\right)....\left(5^{256}+1\right)\)

....

\(E=\left(5^{128}-1\right)\left(5^{128}+1\right)\left(5^{256}+1\right)\)

\(E=\left(5^{256}-1\right)\left(5^{256}+1\right)\)

\(E=5^{512}-1\)