某等差数列,lim(n接近无限大)Sn/(an)²=1/6,求公差.
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某等差数列,lim(n接近无限大)Sn/(an)²=1/6,求公差.
设公差为d,则
an=a1+(n-1)d=dn+a1-d
Sn=na1+n(n-1)d/2
∴Sn/a²n=[na1+n(n-1)d/2]/(dn+a1-d)²
∴lim(n-->∞)Sn/(an)²
=lim(n-->∞)[na1+n(n-1)d/2]/(dn+a1-d)²
=lim(n-->∞)[na1+n(n-1)d/2]/(d²n²+2d(a1-d)n+(a1-d)²]
=lim(n-->∞)[a1/n+(1-1/n)d/2]/(d²+2d(a1-d)/n+(a1-d)²/n²]
=(d/2)/d²=1/(2d)
∵lim(n接近无限大)Sn/(an)²=1/6
∴1/(2d)=1/6 ∴d=3
an=a1+(n-1)d=dn+a1-d
Sn=na1+n(n-1)d/2
∴Sn/a²n=[na1+n(n-1)d/2]/(dn+a1-d)²
∴lim(n-->∞)Sn/(an)²
=lim(n-->∞)[na1+n(n-1)d/2]/(dn+a1-d)²
=lim(n-->∞)[na1+n(n-1)d/2]/(d²n²+2d(a1-d)n+(a1-d)²]
=lim(n-->∞)[a1/n+(1-1/n)d/2]/(d²+2d(a1-d)/n+(a1-d)²/n²]
=(d/2)/d²=1/(2d)
∵lim(n接近无限大)Sn/(an)²=1/6
∴1/(2d)=1/6 ∴d=3
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