Dãy số a \((a_n)\) có số hạng tổng quát \(a_n=1+2+...+n\).
Tổng \(S_n=a_1+a_2+a_3+.....+a_n\) bằng
\(\dfrac{n\left(n+1\right)\left(n+2\right)}{6}\).\(\dfrac{n\left(2n+1\right)\left(n+2\right)}{6}\).\(\dfrac{n\left(n+2\right)\left(2n+3\right)}{6}\).\(\dfrac{\left(n+2\right)\left(n+3\right)n}{6}\).Hướng dẫn giải:
\(a_n=\frac{n\left(n+1\right)}{2}=\frac{n^2+n}{2}\)
=> \(a_1=\frac{1^2+1}{2}\)
\(a_2=\frac{2^2+2}{2}\)
\(a_3=\frac{3^2+3}{2}\)
.......
\(a_n=\frac{n^2+n}{2}\)
=> \(a_1+a_2+....+a_n=\dfrac{\left(1^2+2^2+3^2+......+n^2\right)+\left(1+2+...+n\right)}{2}\)
\(=\dfrac{\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}+\dfrac{n\left(n+1\right)}{2}}{2}=\dfrac{\dfrac{n\left(n+1\right)}{2}\left(\dfrac{2n+1}{3}+1\right)}{2}\)
\(=\dfrac{\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}+\dfrac{n\left(n+1\right)}{2}}{2}=\dfrac{\dfrac{n\left(n+1\right)}{2}\left(\dfrac{2n+1}{3}+1\right)}{2}\)
\(=\dfrac{n\left(n+1\right)\left(n+2\right)}{6}\)
