Question

Tính biểu thức sau một cách Hợp lý

\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

Answer 1

Lời giải:

Áp dụng HĐT đáng nhớ \((a-b)(a+b)=a^2-b^2\). Ta có:

\(A=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)

\(2A=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)

\(=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)

\(=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)

\(=(3^8-1)(3^8+1)(3^{16}+1)(3^{32}+1)\)

\(=(3^{16}-1)(3^{16}+1)(3^{32}+1)\)

\(=(3^{32}-1)(3^{32}+1)=3^{64}-1\)

\(\Rightarrow A=\frac{3^{64}-1}{2}\)

Answer 2

Lời giải:

Áp dụng HĐT đáng nhớ \((a-b)(a+b)=a^2-b^2\). Ta có:

\(A=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)

\(2A=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)

\(=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)

\(=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)

\(=(3^8-1)(3^8+1)(3^{16}+1)(3^{32}+1)\)

\(=(3^{16}-1)(3^{16}+1)(3^{32}+1)\)

\(=(3^{32}-1)(3^{32}+1)=3^{64}-1\)

\(\Rightarrow A=\frac{3^{64}-1}{2}\)