求数列1/2,3/4,5/8,16/17…2n-1/2^n的前n项和Sn = 1×1/2 + 3×1/4 + 5×1/8 + ...+ (2n-3)/2^(n-1) + (2n-1)/2^nSn/2 = 1×1/4 + 3×1/8 + 5×1/16 + ...+ (2n-3)/2^n + 2n-1/2^(n+1)Sn - Sn/2 = 1/2 - 1 + 2( 1/2 + 1/4 + 1/8 + ...+ 1/2^n ) - 2n-1/2^(n+1)S
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求数列1/2,3/4,5/8,16/17…2n-1/2^n的前n项和Sn = 1×1/2 + 3×1/4 + 5×1/8 + ...+ (2n-3)/2^(n-1) + (2n-1)/2^nSn/2 = 1×1/4 + 3×1/8 + 5×1/16 + ...+ (2n-3)/2^n + 2n-1/2^(n+1)Sn - Sn/2 = 1/2 - 1 + 2( 1/2 + 1/4 + 1/8 + ...+ 1/2^n ) - 2n-1/2^(n+1)S
求数列1/2,3/4,5/8,16/17…2n-1/2^n的前n项和
Sn = 1×1/2 + 3×1/4 + 5×1/8 + ...+ (2n-3)/2^(n-1) + (2n-1)/2^n
Sn/2 = 1×1/4 + 3×1/8 + 5×1/16 + ...+ (2n-3)/2^n + 2n-1/2^(n+1)
Sn - Sn/2 = 1/2 - 1 + 2( 1/2 + 1/4 + 1/8 + ...+ 1/2^n ) - 2n-1/2^(n+1)
Sn = -1/2 + 2×(1/2)×(1-1/2^n)/(1-1/2) - 2n-1/2^(n+1)
= 3 - (2n+3)/2^n
为什么第三步Sn - Sn/2 = 1/2 - 1 ...这个 -1 是哪里来的?
求数列1/2,3/4,5/8,16/17…2n-1/2^n的前n项和Sn = 1×1/2 + 3×1/4 + 5×1/8 + ...+ (2n-3)/2^(n-1) + (2n-1)/2^nSn/2 = 1×1/4 + 3×1/8 + 5×1/16 + ...+ (2n-3)/2^n + 2n-1/2^(n+1)Sn - Sn/2 = 1/2 - 1 + 2( 1/2 + 1/4 + 1/8 + ...+ 1/2^n ) - 2n-1/2^(n+1)S
因为2( 1/2 + 1/4 + 1/8 + ...+ 1/2^n ) 中的1/2是凑出来的,原题没,
为保证原式值不变,要减1
2( 1/2 + 1/4 + 1/8 + ...+ 1/2^n )，这部分被翻倍了
所以要减回去，1/2的n次方求和就是1，当然多了个尾项，自己斟酌
根据前面的公式来Sn=1/2所以-Sn/2=-(1/2)/2=-1