2个回答
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奇数项是以a1为首项,2d为公差的等差数列,共n+1项。偶数项是以a2为首项,2d为公差的等差数列,共n项。
所求值=[(n+1)a1+(1+2+...+n)(2d)]/[n(a1+d)+(1+2+...+n-1)(2d)]
=[(n+1)a1+n(n+1)d]/[na1+nd+(n-1)nd]
=[(n+1)(a1+nd)]/(na1+nd+n²d-nd)
=[(n+1)(a1+nd)]/(na1+n²d)
=[(n+1)(a1+nd)]/[n(a1+nd)]
=(n+1)/n
选B
所求值=[(n+1)a1+(1+2+...+n)(2d)]/[n(a1+d)+(1+2+...+n-1)(2d)]
=[(n+1)a1+n(n+1)d]/[na1+nd+(n-1)nd]
=[(n+1)(a1+nd)]/(na1+nd+n²d-nd)
=[(n+1)(a1+nd)]/(na1+n²d)
=[(n+1)(a1+nd)]/[n(a1+nd)]
=(n+1)/n
选B
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