两道数学题(1+1/2)×(1-1/2)×(1+1/3)×(1-1/3)×…×(1+1/99)×(1-1/99)注:1/

3个回答

  • 1

    原式=(1/2)*(3/2)*(2/3)*(4/3)*(3/4)*(5/4)*.(98/99)*

    (100/99)

    =分子与分母相同的就相抵消

    =(1/2)*(100/99)

    =50/99

    2

    1/2+3/4+7/8+15/16+31/32+63/64+127/128+255/256

    =(1-1/2)+(1-1/4)+(1-1/8).(1-1/256)

    =(1+1+1+1+1+1+1+1)-(1/2+1/4+1/8+1/16+1/32+.1/256)

    =8-(1-1/256)

    =8-1+1/256

    =7+1/256

    另解

    an=(2^n-1)/2^n=1-1/2^n

    Sn=8*1-(1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256)

    =8-0.5(1-0.5^8)/1-0.5

    =8-255/256

    =1793/256

    3

    (1/2×3×4)+(1/3×4×5)+(1/4×5×6)+…+(1/8×9×10)

    =6+20/3+15/2+42/5+28/3+72/7+45/4

    =8321/140