Question

calculați. (4+8+12+16+....+4000)÷(2+4+6+8+....+2000)+(2011+37)÷2^10=​

Answer 1

 

$\displaystyle\bf\\(4+8+12+16+...+4000):(2+4+6+8+...+2000)+(2011+37):2^{10}=\\\\=\frac{4+8+12+16+...+4000)}{2+4+6+8+...+2000}+\frac{2011+37}{2^{10}}=\\\\\\=\frac{4(1+2+3+4+...+1000)}{2(1+2+3+4+...+1000)}+\frac{2048}{2^{10}}=\\\\\\=\frac{4(1+2+3+4+...+1000)}{2(1+2+3+4+...+1000)}+\frac{2048}{2^{10}}=\\\\\\=\frac{4}{2}+\frac{2^{11}}{2^{10}}=\\\\\\=2+2^{11-10}=2+2=\boxed{\bf4}$