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
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/28 04:15:05
![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](/uploads/image/z/5466324-12-4.jpg?t=%E8%AE%BESn%3D1%2B2%2B3%2B%E2%80%A6%E2%80%A6%2Bn%2C%E5%88%99f%28n%29%3DSn%2F%28n%2B32%29Sn%2B1%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E6%98%AF%E5%A4%9A%E5%B0%91Sn%3D1%2B2%2B3%2B%E2%80%A6%E2%80%A6%2Bn%3Dn%28n%2B1%29%2F2S%28n%2B1%29%3D%28n%2B1%29%28n%2B2%29%2F2%3Bf%28n%29%3Dsn%2F%28n%2B32%29s%28n%2B1%29%3D%5Bn%28n%2B1%29%2F2%5D%2F%5B%28n%2B32%29%2A%28n%2B1%29%28n%2B2%29%2F2%5D%3Dn%2F%28n%2B32%29%28n%2B2%29%3Dn%2F%28%28n%5E2%2B34n%2B64%29%3D1%2F%28n%2B64%2Fn%2B34%29%E7%94%B1%E4%BA%8Ex%2B64%2Fx%3E%3D2%E6%A0%B9%E5%8F%B764%3D16+%E6%AD%A4%E6%97%B6x%3D8%E4%B9%9F%E5%B0%B1)
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.