计算1⼀2+2的2次方分之3+2的3次方分之5+2的4次方分之7+...+2的50次方分之99

错位相减法求和：
记S=1/2+3/2^2+5/2^3+7/2^4+...+99/2^50 ①
两边乘以1/2:
1/2S=1/2^2+3/2^3+5/2^4+......+97/2^50+99/2^51 ②
①-②:
1/2S=1/2+2*1/2^2+2*1/2^3+2*1/2^4+.......+2*1/2^50-99/2^51
=1/2+2(1/2^2+1/2^3+1/2^4+.....+1/2^50)-99/2^51
=1/2+2*1/4[1-1/2^49]/(1-1/2)-99/2^51
=3/2-103/2^51
∴S=3-103/2^50

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错位相减法求和：
记S=1/2+3/2^2+5/2^3+7/2^4+...+99/2^50 ①
两边乘以1/2:
1/2S=1/2^2+3/2^3+5/2^4+......+97/2^50+99/2^51 ②
①-②:
1/2S=1/2+2*1/2^2+2*1/2^3+2*1/2^4+.......+2*1/2^50-99/2^51
=1/2+2(1/2^2+1/2^3+1/2^4+.....+1/2^50)-99/2^51
=1/2+2*1/4[1-1/2^49]/(1-1/2)-99/2^51
=3/2-103/2^51
∴S=3-103/2^50
希望帮到你，不懂请追问，满意请采纳

错位相减法求和：
记S=1/2+3/2^2+5/2^3+7/2^4+...+99/2^50 ①
两边乘以1/2:
1/2S=1/2^2+3/2^3+5/2^4+......+97/2^50+99/2^51 ②
①-②:
1/2S=1/2+2*1/2^2+2*1/2^3+2*1/2^4+.......+2*1/2^50-99/2^51
=1/2+2(1/2^2+1/2^3+1/2^4+.....+1/2^50)-99/2^51
=1/2+2*1/4[1-1/2^49]/(1-1/2)-99/2^51
=3/2-103/2^51
∴S=3-103/2^50