等差数列bn=(a1+a2+3a….an)/n
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等差数列bn=(a1+a2+3a….an)/n
(1)bn=n^2,求{an}
(2){bn}为等差数列,求证{an}也为等差数列
错了 是 (a1+a2-+a3...+an)/n
(1)bn=n^2,求{an}
(2){bn}为等差数列,求证{an}也为等差数列
错了 是 (a1+a2-+a3...+an)/n
n=(a1+a2+3a….an)/n=Sn/n
b1=a1
bn=n^2,a1=b1=1
sn=n^3
s(n+1)=(n+1)^3
a(n+1)=s(n+1)-sn=3n^2+3n+1=3n(n+1)+1
所以an=3n(n-1)+1 n>=2
当n=1时,a1=1;
2)
如果bn是等差数列,不妨设bn=kn+d;
则Sn=(kn+d)n
s(n+1)=(kn+k+d)(n+1)
a(n+1)=s(n+1)-sn=(kn+k+d)(n+1)-(kn+d)n
=(kn+d)+k(n+1)=2kn+(k+d)
是等差数列.
{an}也为等差数列
b1=a1
bn=n^2,a1=b1=1
sn=n^3
s(n+1)=(n+1)^3
a(n+1)=s(n+1)-sn=3n^2+3n+1=3n(n+1)+1
所以an=3n(n-1)+1 n>=2
当n=1时,a1=1;
2)
如果bn是等差数列,不妨设bn=kn+d;
则Sn=(kn+d)n
s(n+1)=(kn+k+d)(n+1)
a(n+1)=s(n+1)-sn=(kn+k+d)(n+1)-(kn+d)n
=(kn+d)+k(n+1)=2kn+(k+d)
是等差数列.
{an}也为等差数列
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