设数列的前n项和为sn=n^2-4n+1,则/ai/+/a2/+/a3/+.+/a10/=?

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设数列的前n项和为sn=n^2-4n+1,则/ai/+/a2/+/a3/+.+/a10/=?
设数列的前n项和为sn=n^2-4n+1,
则/ai/+/a2/+/a3/+.+/a10/=?

设数列的前n项和为sn=n^2-4n+1,则/ai/+/a2/+/a3/+.+/a10/=?
Sn = n^2 - 4n + 1
S(n-1) = n^2 - 2n + 1 - 4n + 4 + 1 = n^2 - 6n + 6
An = (n^2 - 4n + 1) - (n^2 - 6n + 6) = 2n-5 (n>1)
S1 = a1 = 1-4+1 = -2
a2 = 4-8+1+2 = -1
a3..a10 = 2n-5 > 0
所以
/a1/+/a2/+/a3/+.+/a10/
= -2a1-2a2+a1+a2+a3+.+a10
= -2a1-2a2+S10
= -2*-2 + -1*-2 + 100-40+1
= 4+2+60-1
= 65

把10直接代到前n项和公式就行。
是不是97

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