an=3n^2-2n+1 求sn=?
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an=3n^2-2n+1 求sn=?
an=3n^2-2n+1 求sn=?
an=3n^2-2n+1 求sn=?
an=bn-cn
bn=3n^2
cn=-2n+1
1)sbn=1²+2²+3²+……+n²=3[1/6·n(n+1)(2n+1)]
证明如下:
不妨设1²+2²+3²+……+n²=S
利用恒等式(n+1)³=n³+3n²+3n+1,得:
(n+1)³-n³=3n²+3n+1
n³-(n-1)³=3(n-1)²+3(n-1)+1
………………………………
3³-2³=3·2²+3·2+1
2³-1³=3·1²+3·1+1
将这n个式子两端分别相加,得:
(n+1)³-1=3(1²+2²+3²+……+n²)+3(1+2+3+……+n)+n
由于1+2+3+4+……+n=n(n+1)/2
代入上式,得:
n³+3n²+3n=3S+3/2×n(n+1)+n
整理后得S=1/6·n(n+1)(2n+1)
即1²+2²+3²+……+n²=1/6·n(n+1)(2n+1)
2)scn=-n(n+1)+n=-n^2
所以sn=sbn+scn=3[1/6·n(n+1)(2n+1)]-n^2
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