Đáp án:
22075
Giải thích các bước giải:
$ \text{Áp dụng đẳng thức:} 1^2+2^2..+n^2=\dfrac{n.(n+1)(2n+1)}{6}\\ A=1.3+3.5+5.7+...+49.51\\ \rightarrow A=1.(1+2)+3(3+2)+...+49(49+2)\\ \rightarrow A=(1^2+3^2+5^2+...+49^2)+2(1+3+5+..+49)\\ \rightarrow A=(1^2+2^2+3^2+4^2+5^2+...+49^2+50^2)-(2^2+4^2+...+50^2)+2(1+3+5+..+49)\\ \rightarrow A=(1^2+2^2+3^2+4^2+5^2+...+49^2+50^2)-4(1^2+2^2+...+25^2)+2(1+3+5+..+49)\\ \rightarrow A=\dfrac{50(50+1)(2.50+1)}{6}-4.\dfrac{25(25+1)(2.25+1)}{6}+2.25.(49+1)/2\\ \rightarrow A=\dfrac{50(50+1)(2.50+1)}{6}-4.\dfrac{25(25+1)(2.25+1)}{6}+2.25.(49+1)/2\\ \rightarrow A=22075$
