a(n+1)=a1+nd
an=a1+(n-1)d
a(n+1)+an=2a1+(2n-1)d=4n-3
整理,得
2dn+2a1-d=4n-3
2d=4
2a1-d=-3
解得
d=2 a1=-1/2
a1=-1/2
已知等差數(shù)列an+1+an=4n-3求a1
已知等差數(shù)列an+1+an=4n-3求a1
①當(dāng)a1=2時(shí),求Sn
①當(dāng)a1=2時(shí),求Sn
數(shù)學(xué)人氣:572 ℃時(shí)間:2020-02-05 15:22:02
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