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七(下)一道数学题小刚在计算3(4+1)(4^2+1)时,把3写作4-1后,得3(4+1)(4^2+1)=(4-1)(4

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七(下)一道数学题
小刚在计算3(4+1)(4^2+1)时,把3写作4-1后,得
3(4+1)(4^2+1)=(4-1)(4+1)(4^2+1)=(4^2-1)(4^2+1)=4^4-1
阅读上面的材料后解答下面问题:
(1)求(2+1)(2^2+1)(2^4+1)(2^8+1)...(2^64+1)的值;
(2)求(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)的值
最好用平方差公式!详细解答的有+50分的机会(不要写得密密麻麻)
1.
(2+1)(2^2+1)(2^4+1)(2^8+1)...(2^64+1)
=1 *(2+1)(2^2+1)(2^4+1)(2^8+1)...(2^64+1)
=(2-1)*(2+1)(2^2+1)(2^4+1)(2^8+1)...(2^64+1)
=...
=(2^64-1)(2^64+1)
=2^128-1
2.(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)
=[(1-1/2)(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)]/[1/2]
=...
=[(1-1/2^8)(1+1/2^8)]/[1/2]
=[1-1/2^16]/[1/2]
=2[1-1/2^16]
=2-1/2^15