ak≥a(k-1),且ak≥a(k+1)
∴k(k+4)*(2/3)^k≥(k-1)(k+3)*(2/3)^(k-1)①
k(k+4)*(2/3)^k≥(k+1)(k+5)*(2/3)^(k+1)②
①==> 2k(k+4)≥3(k-1)(k+3)
==> k²-2k-9≤0
==> (k-1)²≤10
==>1-√10≤k≤1+√10
k≥1 ∴k≤4
②==>3k(k+4)≥2(k+1)(k+5)
==> k²≥10
==>k≥√10
==>k≥4
∴k=4
這樣吧,不懂之處追問(wèn)或Hi我