数列1/(1^2+2),1/(2^2+4),1/(3^2+6),1/(4^2+8),......1/(n^2+2n),......前n项和为多少?

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数列1/(1^2+2),1/(2^2+4),1/(3^2+6),1/(4^2+8),......1/(n^2+2n),......前n项和为多少?

数列1/(1^2+2),1/(2^2+4),1/(3^2+6),1/(4^2+8),......1/(n^2+2n),......前n项和为多少?
数列1/(1^2+2),1/(2^2+4),1/(3^2+6),1/(4^2+8),......1/(n^2+2n),......前n项和为多少?

数列1/(1^2+2),1/(2^2+4),1/(3^2+6),1/(4^2+8),......1/(n^2+2n),......前n项和为多少?
看通项:分母可分成:n*(n+2)
则有:1/(n^2+2n)=(1/2)*(1/n-1/(n+2))
中间的项全消掉,则有:(1/2)*(1-1/(n+2))

1/(n^2+2n)=0.5*[1/n-1/(n+2)]
故裂项求和前n项和为0.5*[1-1/(n+2)]

好难啊

看通项:分母可分成:n*(n+2)
则有:1/(n^2+2n)=(1/2)*(1/n-1/(n+2))
中间的项全消掉,则有:(1/2)*(1-1/(n+2))