\(A=\frac{5}{2^2}+\frac{10}{3^2}+\frac{17}{4^2}+...+\frac{226}{15^2}=\frac{2^2+1}{2^2}+\frac{3^2+1}{3^2}+\frac{4^2+1}{4^2}+.+\frac{15^2+1}{15^2}.\)
Vì A có 14 số hạng nên : \(A=1+\frac{1}{2^2}+1+\frac{1}{3^2}+1+\frac{1}{4^2}+...+1+\frac{1}{15^2}=14+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{15^2}.\)
\(\Rightarrow A< 14+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{14.15}=14+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...\frac{1}{14}-\frac{1}{15}.\)
\(\Rightarrow A=15-\frac{1}{15}< 15.\) Lạy có :
\(A>14+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{15.16}=14+\frac{1}{2}-\frac{1}{16}< 14,5.\)
Vậy A không phải là số tự nhiên \(14,5< A< 15.\)
