\(A=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{16}\left(1+..+16\right)\\ =1+\dfrac{1}{2}.\dfrac{\left(1+2\right).2}{2}+\dfrac{1}{3}.\dfrac{\left(1+3\right).3}{2}+....+\dfrac{1}{16}.\dfrac{\left(16+1\right).16}{2}\\ =\dfrac{2}{2}+\dfrac{\left(1+2\right)}{2}+...+\dfrac{16+1}{2}\\ =\dfrac{2+...+17}{2}\\ =\dfrac{152}{2}=76\)
Ủng hộ 1 cách khác :v
Xét thừa số tổng quát: \(\dfrac{1}{n}\left(1+2+3+...+n\right)=\dfrac{1+2+3+...+n}{n}=\dfrac{\left[\left(n-1\right):1+1\right]:2\left(n+1\right)}{n}=\dfrac{\dfrac{n}{2}\left(n+1\right)}{n}=\dfrac{\dfrac{n^2}{2}+\dfrac{n}{2}}{n}=\dfrac{n\left(\dfrac{n}{2}+\dfrac{1}{2}\right)}{n}=\dfrac{n+1}{2}\)
Hay:
\(A=\dfrac{2}{2}+\dfrac{2+1}{2}+\dfrac{3+1}{2}+...+\dfrac{16+1}{2}\)
\(A=\dfrac{2+3+4+...+17}{2}\)
Đơn giản r :v
