1/1x2x3+1/2x3x4+1/3x4x5+.+1/n(n+1)(n+2)
1/1x2x3+1/2x3x4+1/3x4x5+.+1/n(n+1)(n+2)
1/1x2x3+1/2x3x4+1/3x4x5+.+1/n(n+1)(n+2)
1/1x2x3+1/2x3x4+1/3x4x5+.+1/n(n+1)(n+2)
Sn=1/[n(n+1)(n+2)]=(1/2){1/[n)n+1)]-1/[(n+1)(n+2)]}
=(1/2)[1/n-1/(n+1)-1/(n+1)+1/(n+2)]
=(1/2)[1/n-2/(n+1)+1/(n+2)]
=1/1x2x3+1/2x3x4+1/3x4x5+...+1x/n(n+1)(n+2)
=(1/2)[1/1-2/2+1/3+1/2-2/3+1/4+1/3-2/4+1/5+/4-2/5+1/6
+.+1/n-2/(n+1)+1/(n+2)]
=(1/2)[1-1/2-1/(n+1)+1/(n+2)]
=(n^2+3n)/[4(n+1)(n+2)]
1/1x2x3+1/2x3x4+1/3x4x5+...........+1/n(n+1)(n+2)
=1/2×[1/1x2-1/2x3+1/2x3-1/3x4+1/3x4-1/4x5+...........+1/n(n+1)-1/(n+1)(n+2)]
=1/2×[1/1x2-1/(n+1)(n+2)]
=1/2×[1/2-1/(n+1)(n+2)]
=1/2×(n²+3n)/2(n+1)(n+2)
=(n²+3n)/4(n+1)(n+2)
这个是求极限的题 原题可以化为 1(1/2-1/3)+1/2(1/3-1/4)+1/3(1/4-1/ 5)+....+1/n(1/n+1*1/n+2) = 1/2-1/1*2*3-1/2*3*4-1/3*4*5-...-1/n*(n-1)*(n+1)-1/n*1/(n+2)
将后一个式子与原式相加 得值为 1/2 - 1/n*1/(n+2)
n 无限大 ...
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这个是求极限的题 原题可以化为 1(1/2-1/3)+1/2(1/3-1/4)+1/3(1/4-1/ 5)+....+1/n(1/n+1*1/n+2) = 1/2-1/1*2*3-1/2*3*4-1/3*4*5-...-1/n*(n-1)*(n+1)-1/n*1/(n+2)
将后一个式子与原式相加 得值为 1/2 - 1/n*1/(n+2)
n 无限大 1/n*(n+2) 趋于零
则原问题极限值为 1/4
望采纳 采纳
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