等差数列{an}中,an=n,设cn=2n-1/2n,Tn=c1+c2+c3...+cn 求证:Tn>-1/2根号n
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等差数列{an}中,an=n,设cn=2n-1/2n,Tn=c1+c2+c3...+cn 求证:Tn>-1/2根号n
...这题和an有什么关系吗?
cn=(2n-1)/2n,Tn=c1+c2+...+cn.
当n=1时,Tn=1/2>-1/2成立.
假设当n=k时也成立,即c1+c2+...+ck>-1/2√(k)①
则n=k+1时,①两边同时加上c(k+1).
c1+..ck+(2k+1)/(2k+2)>-1/2√(k)+(2k+1)/(2k+2)
要证T(k+1)>-1/2√(k+1)
只需-1/2√(k)+(2k+1)/(2k+2)≥-1/2√(k+1)
只需1/2√(k+1)-1/2√(k)+(2k+1)/(2k+2)≥0
只需1/2(√(k+1)-√(k))+(2k+1)/(2k+2)≥0
1/2(√(k+1)-√(k))>0,(2k+1)/(2k+2)>0,.
∴n=k+1时结论也成立.
综上,命题成立.
cn=(2n-1)/2n,Tn=c1+c2+...+cn.
当n=1时,Tn=1/2>-1/2成立.
假设当n=k时也成立,即c1+c2+...+ck>-1/2√(k)①
则n=k+1时,①两边同时加上c(k+1).
c1+..ck+(2k+1)/(2k+2)>-1/2√(k)+(2k+1)/(2k+2)
要证T(k+1)>-1/2√(k+1)
只需-1/2√(k)+(2k+1)/(2k+2)≥-1/2√(k+1)
只需1/2√(k+1)-1/2√(k)+(2k+1)/(2k+2)≥0
只需1/2(√(k+1)-√(k))+(2k+1)/(2k+2)≥0
1/2(√(k+1)-√(k))>0,(2k+1)/(2k+2)>0,.
∴n=k+1时结论也成立.
综上,命题成立.
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