求解(3+1)(3^2+1)(3^4+1).(3^2008+1)-3^4016/2
1个回答
(3-1)*原式=(3-1)(3+1)(3^2+1)(3^4+1)……(3^2008+1)-3^4016=3^4016-1-3^4016=-1
原式=-1/2
相关问题
(3+1)(3^2+1)(3^4+1)…(3^2008+1)-3^4016/2
计算:(3+1)(3^2+1)(3^4+1)…(3^2008+1)-3^4016/2
1/3+1/(1+2)*(3+2)+1/(1+4)*(3+4)+.1/(1+2008)*(3+2008)
1/(2/3)/(3/4)/(4/5)/./(4014/4015)/(4015/4016)=?
1/1*2+1/2*3+1/3*4+...+1/2007*2008+1/2008*2009
1/2*1+1/3*2+1/4*3.+1/2009*2008=
1/1*2+1/2*3+1/3*4+........+1/2007*2008
1/(1*2)+1/(2*3)+1/(3*4)+.1/(2007*2008)=
1,计算(6×4014+9×4016+2/1)÷(3×4014+3×6024+4/1) 2,
求{1/(1+√2)+1/(√2+√3)+1/(√3+√4)+...+1/(√2008+√2009)