sn=2n^2-n,bn=sn/(n+p)=(2n^2-n)/(n+p)
b1=1/(1+p),b2=6/(2+p),b3=15/(3+p).
bn是等差数列,则b1+b3=2b2,即1/(1+p)+15/(3+p)=12/(p+2),通分,解得p=0
已知数列an前N项和为sn,点(n,sn)都在函数f(x)=2x^2-x上,设bn=sn/(n+p),且数列bn是等差数列,求P
已知数列an前N项和为sn,点(n,sn)都在函数f(x)=2x^2-x上,设bn=sn/(n+p),且数列bn是等差数列,求P
p为非零数额
p为非零数额
数学人气:783 ℃时间:2019-12-03 07:35:55
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