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求数列2又1/4,4又1/8,6又1/16,...,2n+1/2^(n+1),...的前n项的和Sn
人气:211 ℃ 时间:2020-03-27 13:15:55
解答
2又1/4+4又1/8+6又1/16+,...,+2n+1/2^(n+1)
=(2+4+6+……+2n)+[1/4+1/8+1/16+……+1/2^(n+1)]
=(2+2n)n/2+1/2-1/2^(n+1)
=n(n+1)+1/2-1/2^(n+1)
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