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求数列{n/2^n}的前n项和
2^n表示2的n次方.
Sn = 1/2 + 2/2^2 + 3/2^3 + ... + n/2^n
2Sn = 1 + 2/2^1 + 3/2^2 + ... + n/2^(n-1)
所以 Sn
= 2Sn - Sn
= (1 + 2/2 + 3/2^2 + ... + n/2^(n-1)) - (1/2 + 2/2^2 + 3/2^3 + ... + n/2^n)
= 1 + 1/2 + 1/2^2 + ... + 1/2^(n-1) - n/2^n
= 2 - 1/2^(n-1) - n/2^n
= 2 - (n+2)/2^n
2^n表示2的n次方.
Sn = 1/2 + 2/2^2 + 3/2^3 + ... + n/2^n
2Sn = 1 + 2/2^1 + 3/2^2 + ... + n/2^(n-1)
所以 Sn
= 2Sn - Sn
= (1 + 2/2 + 3/2^2 + ... + n/2^(n-1)) - (1/2 + 2/2^2 + 3/2^3 + ... + n/2^n)
= 1 + 1/2 + 1/2^2 + ... + 1/2^(n-1) - n/2^n
= 2 - 1/2^(n-1) - n/2^n
= 2 - (n+2)/2^n
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