在等差数列中,若Sm/Sn=m^2/n^2(m不等于n),则am/an=答案是(2m-1)/(2n-1)要详细的过程,谢谢!
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![在等差数列中,若Sm/Sn=m^2/n^2(m不等于n),则am/an=答案是(2m-1)/(2n-1)要详细的过程,谢谢!](/uploads/image/z/6162252-60-2.jpg?t=%E5%9C%A8%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E4%B8%AD%2C%E8%8B%A5Sm%2FSn%3Dm%5E2%2Fn%5E2%28m%E4%B8%8D%E7%AD%89%E4%BA%8En%29%2C%E5%88%99am%2Fan%3D%E7%AD%94%E6%A1%88%E6%98%AF%EF%BC%882m-1%29%2F%282n-1%29%E8%A6%81%E8%AF%A6%E7%BB%86%E7%9A%84%E8%BF%87%E7%A8%8B%EF%BC%8C%E8%B0%A2%E8%B0%A2%EF%BC%81)
在等差数列中,若Sm/Sn=m^2/n^2(m不等于n),则am/an=答案是(2m-1)/(2n-1)要详细的过程,谢谢!
在等差数列中,若Sm/Sn=m^2/n^2(m不等于n),则am/an=
答案是(2m-1)/(2n-1)
要详细的过程,谢谢!
在等差数列中,若Sm/Sn=m^2/n^2(m不等于n),则am/an=答案是(2m-1)/(2n-1)要详细的过程,谢谢!
原问题即 :有两个数列,{An} {Bn},若 Sm :Sn = m^2 :n^2
求 Am :Bn (公差分别为d1,d2)
Sm = A1d + 0.5*m(m-1)d1 = 0.5d1m^2 + m(A1-0.5d1)
Sn = 0.5d2n^2 + n(B1-0.5d2)
∵ Sm :Sn = m^2 :n^2 ∴ A1-0.5d1=0 B1-0.5d2=0
Sm = 0.5d1m^2 Sn = 0.5d2n^2 ∴d1=d2=d A1=A2=0.5d
∵ Am = A1 + (m-1)d = d(m-0.5) Bn = d(n-0.5)
∴ Am :Bn = (m-0.5) :(n-0.5) = (2m-1) :(2n-1)
Let T_m=S_m/m^2.
Then T_m=T_n , for all m,n
So T_m=T_1=S_1=a_1, that is, S_m=a_1*m^2.
Hence a_m=a_1(m^2-(m-1)^2)=a_1(2m-1).
Thus we have a_m/a_n=(2m-1)/(2n-1)