Question

1+5+9+.....+(4n-3)=2n la a doua -n ??????

Answer 1

   
[tex]\displaystyle \\ \texttt{Sirul are n termeni.} \\ \texttt{Aplicam formula lui GAUSS:} \\ \\ 1+5+9+.....+(4n-3) = \frac{\text{(numarul de termeni)(ultimul + primul) }}{2} \\ \\ 1+5+9+.....+(4n-3)= \frac{n(4n-3+1)}{2}= \\ \\ = \frac{n(4n-2)}{2}= \frac{n[2(2n-1)]}{2}= \frac{2n(2n-1)}{2}=n(2n-1)=\boxed{2n^2 -n}\\ \\ [/tex]

Answer 2

$\displaystyle 1+5+9+...+(4n-3)=\sum\limits_{k=1}^n (4k-3)=\sum\limits_{k=1}^n4k-\sum\limits_{k=1}^n3= \\ \\ =4\sum\limits_{k=1}^nk-\sum\limits_{k=1}^n3=\not4 \cdot \frac{n(n+1)}{\not2} -3n=2 n(n+1)-3n= \\ \\ =2n^2+2n-3n=\boxed{2n^2-n}$