已知等差数列{an}的通项公式an=3n-5,求数列的前12项和与前2n-1项的和

2025-03-11 06:33:23
推荐回答(3个)
回答1:

a12=31
a1=-2
S12=(a1+a12)×12÷2=174

a(2n-1)=6n-8
所以S(2n-1)
=(-2+6n-8)(2n-1)/2
=(3n-5)(2n-1)
=6n²-13n+5

回答2:

a1=3*1-5=-2
d=a2-a1=3
Sn=na1+n(n-1)*3/2=-2n+3n(n-1)/2
S12=-24+3*12*(12-1)/2=174
S2n-1=-2*(2n-1)+(2n-1)*(2n-2)*3/2=6n^2-13n+5

回答3:

a1=-2,a12=31,a2n-1=6n-8
S12=12(-2+31)/2=174
S2n-1=(2n-1)(-2+6n-8)/2=(2n-1)(3n-5)
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