Question

ajutor pentru 1+3+5+7+... +8815= Repede va rog

Answer 1

1+3+5+7+...+8815=.... Este de forma 1+3+5+7+...+(2n-1)=n^2; 
8815=2n-1 de unde 2n=8815+1=8816 => n=4408 .Deci suma S=4408^2=19430464
Tine minte urmatoarea formula:
1+3+5+7+....+2n-1=n^2
Demonstratie:
Pentru a calcula suma primelor n numere impare , putem proceda astfel: 1+3+5+7+..........+(2n-1)=(2·0+1)+(2·1+1)+(2·2+1)+(2·3+1)+..............+[2∙(n-1)+1]=
=n+2∙[0+1+2+.........+(n-1)]=n+2∙( n-1)∙n:2=
=n+(n -1)∙n=n∙[1+(n-1)]=n^2

Answer 2

$\displaystyle 1+3+5+7+...+8815 \\ 8815= 1+(n-1) \cdot 2 \\ 8815=1+2n-2 \\ 2n=8815-1+2 \\ 2n= 8816 \\ n=8816:2 \\ n=4408 \\ S_{4408}= \frac{2+4407 \cdot 2}{2} \cdot 4408 \\ \\ S_{4408}= \frac{2+8814}{2} \cdot 4408 \\ \\ S_{4408}= \frac{8816}{2} \cdot 4408 \\ \\ S_{4408}=4408 \cdot 4408 \\ S_{4408}=19430464$