Question

[BTS - Bang Tan Boys]

Cho \(A=\frac{3^3}{1}-\frac{5^3}{3}+\frac{7^3}{6}-\frac{9^3}{10}+\frac{11^3}{15}-\frac{13^3}{21}+\frac{15^3}{28}-\frac{17^3}{36}+...+\frac{199^3}{4950}\)

So sánh \(A\) với 814.

Answer 1

[Nguyễn Việt Lâm]

Sử dụng khá nhiều kiến thức hằng đẳng thức lớp 8, lớp 7 bó tay

\(\frac{A}{2}=\frac{3^3}{2}-\frac{5^3}{6}+\frac{7^3}{12}-\frac{9^3}{20}+...-\frac{197^3}{9702}+\frac{199^3}{9900}\)

\(\frac{A}{2}=\frac{3^3}{1.2}-\frac{5^3}{2.3}+\frac{7^3}{3.4}-\frac{9^3}{4.5}+...+\frac{199^3}{99.100}\)

\(\frac{A}{2}=3^3\left(1-\frac{1}{2}\right)-5^3\left(\frac{1}{2}-\frac{1}{3}\right)+7^3\left(\frac{1}{3}-\frac{1}{4}\right)-...+199^3\left(\frac{1}{99}-\frac{1}{100}\right)\)

\(\frac{A}{2}=3^3-\frac{3^3+5^3}{2}+\frac{5^3+7^3}{3}-\frac{7^3+9^3}{4}+...+\frac{197^3+199^3}{99}-\frac{199^3}{100}\)

\(\frac{A}{2}=3^3-\frac{199^3}{100}-\left(16.2^2+12\right)+\left(16.3^2+12\right)-\left(16.4^2+12\right)+...+\left(16.99^2+12\right)\)

\(\frac{A}{2}=3^3-\frac{199^3}{100}+16\left(3^2-2^2+5^2-4^2+7^2-6^2+...+99^2-98^2\right)\)

\(\frac{A}{2}=3^3-\frac{199^3}{100}+16\left(2+3+4+5+...+98+99\right)\)

\(\frac{A}{2}=3^3-\frac{199^3}{100}+16\left(99.50-1\right)\)

\(\Rightarrow A=16.99.100-\frac{199^3}{50}+22\) (đến đây bấm máy ra kết quả so sánh cũng được)

\(\Rightarrow A=\frac{2^3.100^2\left(100-1\right)-199^3}{50}+22\)

\(A=\frac{200^3-199^3-2.200^2}{50}+22\)

\(A=\frac{200^2+200.199+199^2-2.200^2}{50}+22\)

\(A=\frac{199^2-200^2+200.199}{50}+22\)

\(A=\frac{-199-200+200.199}{50}+22=\frac{199^2}{50}+18\)

\(A< \frac{199.200}{50}+18=814\)

Vậy \(A< 814\)