Question

Sa se arate ca numarul N=8000 supra 19 ori (1supra1+2+...+100 + 1supra1+2+...+101 +...+ 1supra1+2+...+1999) este cub perfect

Answer 1

$N=\dfrac{8000}{19}\cdot\left(\dfrac{1}{1+2+\ldots+100}+\dfrac{1}{1+2+\ldots+101}+\ldots+\dfrac{1}{1+2+\ldots+1999}\right)=\\\\=\dfrac{8000}{19}\cdot\left[\dfrac{1}{\dfrac{(1+100)\cdot 100}{2}}+\dfrac{1}{\dfrac{(1+101)\cdot 101}{2}}+\ldots+\dfrac{1}{\dfrac{(1+1999)\cdot 1999}{2}}\right]=\\\\=\dfrac{8000}{19}\cdot\left(\dfrac{2}{100\cdot 101}+\dfrac{2}{101\cdot 102}+\ldots+\dfrac{2}{1998\cdot 1999}+\dfrac{2}{1999\cdot 2000}\right)=\\\\=\dfrac{16000}{19}\cdot\left(\dfrac{1}{100}-\dfrac{1}{101}+\dfrac{1}{101}-\dfrac{1}{102}+\ldots+\dfrac{1}{1998}-\dfrac{1}{1999}+\dfrac{1}{1999}-\dfrac{1}{2000}\right)=\\\\=\dfrac{16000}{19}\cdot\left(\dfrac{1}{100}-\dfrac{1}{2000}\right)=\dfrac{16000}{19}\cdot\dfrac{19}{2000}=8=2^3,\;care\;este\;cub\;perfect.$

Green eyes.