Sum: SN = (1 / 2) + (3 / 2 ^ 2) + (5 / 2 ^ 3) +... + [(2n-1) / 2 ^ n] It seems to be a process of seeking hope by subtracting dislocation

Sum: SN = (1 / 2) + (3 / 2 ^ 2) + (5 / 2 ^ 3) +... + [(2n-1) / 2 ^ n] It seems to be a process of seeking hope by subtracting dislocation

Yes, your method is correct: 2 * Sn = 1 + (3 / 2) + (5 / 2 ^ 2) +... + [(2n-1) / 2 ^ (n-1)] Sn = 2Sn Sn = 1 + 2 / 2 + 2 / 2 ^ 2 +... + 2 / 2 ^ (n-1) - (2n-1) / 2 ^ n = 1 + 1 + 1 / 2 +... 1 / 2 ^ (n-2) - (2n-1) / 2 ^ n = 1 + (1-1 / 2 ^ (n-1)) * 2 - (2n-1) / 2 ^ n

Sum 1 / 1.3 + 1 / 3.5 + 1 / 5.7 +... + 1 / (2n + 1) (2n-1)

1/（2n+1）（2n-1）
=[（2n+1）-（2n-1）]/【2×（2n+1）（2n-1）】
=1/2×[（2n+1）-（2n-1）]/【（2n+1）（2n-1）】
=1/2×[1/（2n-1）-1/(2n+1)
So 1 / 1 * 3 = 1 / 2 × (1-1 / 3)
1/3*5=1/2×（1/3-1/5）
.
So the original formula = 1 / 2 × (1-1 / 3) + 1 / 2 * (1 / 3-1 / 5) +. + 1 / 2 × [1 / (2n-1) - 1 / (2n + 1)
=1/2*(1-1/3+1/3-1/5+1/5-1/7+.+1/（2n-1）-1/(2n+1))
=1/2*(1-1/(2n+1))
=1/2*2n/(2n+1)
=n/2n+1

Sum: 1-3 + 5-7 + 9-11 + (- 1) ^ n-1 * (2n-1) = how much

If n is odd: the original formula = 1 + (- 3 + 5) + (- 7 + 9) + +（-（2n-3)+（2n-1))=1+2+2+…… +2 (of (n-1) / 2 2) = n if n is even: original formula = (1-3) + (5-7) + +（（2n-3）-（2n-1))=-2-2-2-…… -2 (total N / 2 - 2) = - n to sum up, the original formula = (- 1) ^ (...)