数列求和 i的平方相加（1+4+7+.......n的平方），sn等于什么
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an = (3n-2)^2
= 9n^2-12n+4
=9n(n+1)-21n+4
= 3[n(n+1)(n+2) -(n-1)n(n+1)] - (21/2)[n(n+1) -(n-1)n ] +4
Sn =a1+a2+...+an
=3n(n+1)(n+2) -(21/2)n(n+1) + 4n
= (1/2)n[ 6(n+1)(n+2)- 21(n+1) + 8 ]
= (1/2)n(6n^2-3n -1)
= 9n^2-12n+4
=9n(n+1)-21n+4
= 3[n(n+1)(n+2) -(n-1)n(n+1)] - (21/2)[n(n+1) -(n-1)n ] +4
Sn =a1+a2+...+an
=3n(n+1)(n+2) -(21/2)n(n+1) + 4n
= (1/2)n[ 6(n+1)(n+2)- 21(n+1) + 8 ]
= (1/2)n(6n^2-3n -1)
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