(3-1)×(3+1)×(3^2+1)×(3^4+1)×……×(3^32+1)+2为什么等于(3^64-1)+2
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(1+1/2+1/3+...
1+3+2+1+1x2+3
1+2-3-3-3-3=?
[(2^3-1)(3^3-1)...100^3-1)]/[(2^3+1)(3^3+1)...(100^3+1)]=?
(3+1)(3^2+1)(3^3+1)(3^4+1)…(3^32+1)步骤
1,1,1,2,2,2,3,3,3,填三阶幻方
1/(2^3)+1/(2^3+1)+1/(2^3+2)+1/(2^3+3)+...+1/(2^4-1)
1+1+1+2+1+3+1.100
1,1/2,3/2,3/8,( ),( )
1+1+2+2+3+3+````````````````````````````+99999
求和Sn=1+(1+3)+(1+3+3^2)+(1+3+3^2+3^3)+.+(1+3+3^2+3^3+...+3^n-1)
证明 1+1/1+1/1*2+1/1*2*3+.+1/1*2*3*...*n
1+1/1+2+1/1+2+3+---- +1/1+2+3+ ---- 3Q
(-1/3)^2除以(-1/3)^3*(1/3)^3除以3^-2*(-3)^0
2 1 -2 -3
求证1/1+1/2^(3/2)+1/3^(3/2)…+1/n^(3/2)
计算:2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1
求四阶行列式:2 -1 3 2 3 -3 3 2 3 -1 -1 2 3 -1 3 -1