求数列1/(3n-2)(3n+1)(3n+4)的前n项和
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求数列1/(3n-2)(3n+1)(3n+4)的前n项和
1/(3n-2)(3n+1)-1/(3n+1)(3n+4)=6/(3n-2)(3n+1)(3n+4)
于是1/(3n-2)(3n+1)(3n+4)=(1/6)[1/(3n-2)(3n+1)-1/(3n+1)(3n+4)]
于是1/1*4*7+1/4*7*11+……+1/(3n-2)(3n+1)(3n+4)
=(1/6)[1/1*4-1/4*7+1/4*7-1/7*11+……+1/(3n-2)(3n+1)-1/(3n+1)(3n+4)]
=(1/6)[1/4-1/(3n+1)(3n+4)]
=(3n²+5n)/8(3n+1)(3n+4)【类比1/n(n+1)=1/n-1/(n+1)裂项相消】
于是1/(3n-2)(3n+1)(3n+4)=(1/6)[1/(3n-2)(3n+1)-1/(3n+1)(3n+4)]
于是1/1*4*7+1/4*7*11+……+1/(3n-2)(3n+1)(3n+4)
=(1/6)[1/1*4-1/4*7+1/4*7-1/7*11+……+1/(3n-2)(3n+1)-1/(3n+1)(3n+4)]
=(1/6)[1/4-1/(3n+1)(3n+4)]
=(3n²+5n)/8(3n+1)(3n+4)【类比1/n(n+1)=1/n-1/(n+1)裂项相消】
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