Tổng: \(S=1+3+5+..+\left(2n-3\right)+\left(2n-1\right)\)
Ta có:
\(1+\left(2n-1\right)=1+2n-1=2n\)
\(3+\left(2n-3\right)=3+2n-3=2n\)
\(5+\left(2n-5\right)=5+2n-5=2n\)
.....
\(n+\left(2n-n\right)=n+2n-n=2n\)
Vậy tổng của dãy S là:
\(S=\dfrac{n}{2}\cdot2n=\dfrac{n\cdot2n}{2}=\dfrac{2n^2}{2}=n^2\)