\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{63}+1\right).\)
\(=\frac{\left(3+1\right)\left(3-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{63}+1\right)}{2}\)
\(=\frac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{63}+1\right)}{2}\)
\(=\frac{\left(3^{64}-1\right)\left(3^{63}+1\right)}{2}\left(\text{bn xem lại chỗ }3^{63}\text{ nhé!! ko thì ko lm đc tiếp đâu}\right)\)
ok thế tiếp
\(A=\frac{\left(3^{64}-1\right)\left(3^{64}+1\right)}{2}=\frac{3^{128}-1}{2}.\)
Bài này chỉ áp dụng hằng đẳng thức số 3 thôi ak!
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\left(3-1\right)A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^{64}-1\right)\left(3^{64}+1\right)\)
\(2A=3^{128}-1\)
\(A=\frac{3^{128}-1}{2}\)
