在数列{an}中,an=4n-5/2,a1+a2+···+an=an²+bn,其中a,b为常数,则ab=?why
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在数列{an}中,an=4n-5/2,a1+a2+···+an=an²+bn,其中a,b为常数,则ab=?why
在数列{an}中,an=4n-5/2,a1+a2+···+an=an²+bn,其中a,b为常数,则ab=?
why
在数列{an}中,an=4n-5/2,a1+a2+···+an=an²+bn,其中a,b为常数,则ab=?why
a1+a2+···+an
=4+8+12+.+4n-5/2-5/2-5/2-...-5/2
=(4+4n)*n/2-5n/2
=(2+2n)*n-5n/2
=2n^2+2n-5n/2
=2n^2-n/2
a=2
b=-1/2
ab=-1/2*2=-1
法一:根据题意可得:a1 = 4-5/2 = 3/2
a2 = 8 - 5/2 = 11/2
于是,S1 = a+b = 3/2
S2 = 4a+2b = 7
解此方程组,得:a = 2 b= - 1/2
故,ab = -1
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法一:根据题意可得:a1 = 4-5/2 = 3/2
a2 = 8 - 5/2 = 11/2
于是,S1 = a+b = 3/2
S2 = 4a+2b = 7
解此方程组,得:a = 2 b= - 1/2
故,ab = -1
解法二:根据已知可得:
a1+a2+a3+……+an = 4(1+2+3+……+n) - 5/2*n
= 4*n(n+1)/2 - 5/2n
= 2n^2 - 1/2*n = a*n^2 +bn
据此可见,a = 2 b = -1/2
故:ab = -1
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