求数列1,1+2,1+2+3……的前N项和 求数列1/(1+2)+1/（1+2+3）……+1/（1+2+...
\u6c42\u6570\u52171,1+2,1+2+3\uff0c\u2026\u2026\u7684\u524dn\u9879\u5f97\u548c\u5148\u6c42\u901a\u9879\u516c\u5f0f\u5373Xn
Xn=n\uff081+n\uff09/2=1/2n²+1/2n
\u7136\u540e\u5206\u7ec4\uff0c
Sn=1/2\uff081²+2²+\u2026\u2026+n²\uff09+1/2\uff081+2+\u2026\u2026+n\uff09
=1/2*[1/6 n(n+1)(2n+1)]+1/4 n(n+1)
=1/4 n(n+1)[1/3 (2n+1)+1]
\u7528\u5230\u4e86\u4e00\u4e2a\u516c\u5f0f 1²+2²+\u2026\u2026+n²=1/6 n(n+1)(2n+1)
1
1+2
1+2+3
……
1+2+3+……+N
由于：
1+2+3+……+N=N(N+1)/2=(N²+N)/2
1²+2²+……N²=N(N+1)(2N+1)/6
所以数列各项加起来就是：
S(N)=(1²+1)/2+(2²+2)/2+(3²+3)/2+……+(N²+N)/2
=[(1²+2²+3²+……+N²)+(1+2+3+……+N)]/2
=[N(N+1)(2N+1)/6+N(N+1)/2]/2
=N(N+1)[(2N+1)/6+1/2]/2
=N(N+1)(N+2)/6
每一项都是等差数列求和。第n项是n(n+1)/2，展开后可以看作完全平方数列与等差数列，然后再求和。现将分母变形(1＋2＋3＋…＋n)
变成n(n+1)/2
那么原来的式子=2/(1*2)+2/(2*3)+……+2/n(n+1)列项可得=2*（1-1/n+1)=2n/(n+1)
a(n)=(1+n)*n/2=1/2*(n+n^2)
s(n)=1/2*(1+2+......n）+1/2*（1+4+9+......n^2)
=1/2*(1+n)*n/2+1/2*n(n+1)(2n+1)/6
=n(n+1)(n+2)/6
an=1/2*(n2+n)，sn=1/12(n*(n+1)*(2n+1))+1/4(n*(n+1))
n(n+1)(n+2)
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6
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