如果题目是：2(3+1)(3^2+1)(3^4+1)(3^8+1)+1
则
原式=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)+1
=(3^16-1)+1
=3^16.
2(3+1)(3^2+1)(3^3+1)(3^4+1)(3^8+1)+1
2个回答
相关问题
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2（3+1）（3+1）（3∧4+1）（3∧8+1）…（3∧64+1）+2=
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a1=1/1*2*3+1/2=2/3; a2=1/2*3*4+1/3=3/8,a3=1/3*4*5+1/4=4/15``
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(3+1)(3^2+1)(3^4+1)(3^8+1)=?
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[(1+3)(1+3^2)(1+3^4)(1+3^8).(1+3^32)+1/2]3*3^2*3^3*……*3^10=多
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(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)
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(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)
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1.(3+1)*(3^2+1)*(3^4+1)*(3^8+1)*……*（3^32+1) =
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计算：2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1
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2/3-1/8-1/6= 1/3+2/5+4/15= 1/4+2/5+3/20= 1-3/4-1/8= 4/5-1/3-
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1/1+(1/2-2/1)+(1/3-2/2+3/1)+(1/4-2/3+3/2-4/1）+.+91/9-2/8=3/7