帮忙算几道简算：①(3^/2*4)+(5^/4*6)+(7^/6*8)+(9^/8*10)+(11^

帮忙算几道简算：①(3^/2*4)+(5^/4*6)+(7^/6*8)+(9^/8*10)+(11^

我是真没看明白第一题……2、原式=1/3×（1+1/1+2+1/1+2+3+1/1+2+……+33）=1/3×（2/1×2+2/2×3+……+2/33×34）=2/3×（1-1/34）=11/173、原式=1-1/2+1-1/4+1-1/8+……+1-1/256=8-（1/2+1/4+1/8+……+1/256）=8-（1-1/256）=7又255/256======以下答案可供参考======供参考答案1:①(3^/2*4)+(5^/4*6)+(7^/6*8)+(9^/8*10)+(11^/10*12)=9/8+25/24+49/48+81/80+121/120=1+1/8+1+1/24+1+1/48+1+1/80+1+1/120=5+1/8+1/24+1/48+1/80+1/120=5+1/4×(1/2+1/6+1/12+1/20+1/30) =5+1/4×(1/1*2+1/2*3+1/3*4+1/4*5+1/5*6) =5+1/4×(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6) =5+1/4×(1-1/6) =5+1/4×5/6 =5+5/24=5又5/24②（1/3）+（1/3+6）+（1/3+6+9）+……+（1/3+6+9+……+99）首先，提出1/3得到：1/3*[1+1/(1+2)+1/(1+2+3)+…+1/(1+2+3+…+33)]然后利用等差数列求和公式得到通项公式为：n(n+1)/2则1/3*[1+1/(1+2)+1/(1+2+3)+…+1/(1+2+3+…+33)]=1/3*{2/[1(1+1)]+2/[2(2+1)]+2/[3(3+1)]+…+2/[33(33+1)]}=2/3*{1-1/2+1/2-1/3+…+1/33-1/34}=2/3*{1-1/34}=11/17③1/2+3/4+7/8+15/16+31/32+63/64+127/128+255/256=（1-1/2)+(1-1/4)+(1-1/8)+(1-1/16)+(1-1/32)+(1-1/64)+(1-1/128)+(1-1/256)=1*8-(1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256)=8-(1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/256)+1/256=8-(1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/128)+1/256=8-(1/2+1/4+1/8+1/16+1/32+1/64+1/64)+1/256=8-(1/2+1/4+1/8+1/16+1/32+1/32)+1/256=8-(1/2+1/4+1/8+1/16+1/16)+1/256=8-(1/2+1/4+1/8+1/8)+1/256=8-(1/2+1/4+1/4)+1/256=8-(1/2+1/2)+1/256=8-1+1/256=7+1/256所以，结果为：七又二百五十六分之一供参考答案2:去菁优网看

这下我知道了