设Sn=1+2+3+……+n,则f(n)=Sn/(n+32)Sn+1的最大值是多少Sn=1+2+3+……+n=n(n+1)/2S(n+1)=(n+1)(n+2)/2;f(n)=sn/(n+32)s(n+1)=[n(n+1)/2]/[(n+32)*(n+1)(n+2)/2]=n/(n+32)(n+2)=n/((n^2+34n+64)=1/(n+64/n+34)由于x+64/x>=2根号64=16 此时x=8也就
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设Sn=1+2+3+……+n,则f(n)=Sn/(n+32)Sn+1的最大值是多少Sn=1+2+3+……+n=n(n+1)/2S(n+1)=(n+1)(n+2)/2;f(n)=sn/(n+32)s(n+1)=[n(n+1)/2]/[(n+32)*(n+1)(n+2)/2]=n/(n+32)(n+2)=n/((n^2+34n+64)=1/(n+64/n+34)由于x+64/x>=2根号64=16 此时x=8也就
设Sn=1+2+3+……+n,则f(n)=Sn/(n+32)Sn+1的最大值是多少
Sn=1+2+3+……+n=n(n+1)/2
S(n+1)=(n+1)(n+2)/2;
f(n)=sn/(n+32)s(n+1)=[n(n+1)/2]/[(n+32)*(n+1)(n+2)/2]
=n/(n+32)(n+2)=n/((n^2+34n+64)=1/(n+64/n+34)
由于
x+64/x>=2根号64=16 此时x=8
也就是n=8是n+64/n有最小值16
此时f(n)有最大值1/(16+34)=1/50
帮我解释一下这一步x+64/x>=2根号64=16 此时x=8
设Sn=1+2+3+……+n,则f(n)=Sn/(n+32)Sn+1的最大值是多少Sn=1+2+3+……+n=n(n+1)/2S(n+1)=(n+1)(n+2)/2;f(n)=sn/(n+32)s(n+1)=[n(n+1)/2]/[(n+32)*(n+1)(n+2)/2]=n/(n+32)(n+2)=n/((n^2+34n+64)=1/(n+64/n+34)由于x+64/x>=2根号64=16 此时x=8也就
此步为不等式的性质,两正数a,b的算数平均数(a+b)/2不小于几何平均数根号(ab).
可由(a+b)/2-根号(ab)=(根号a-根号b)^2/2>=0证明.