ajutati ma
calculati urmatoarele sume
1+2+3....+80;
1+3+5....+99;
Răspunsuri la întrebare
\begin{gathered}a)~S=~1+~2+3+~....+~80=\\\\ S= \frac{[80(80+1)]}{2} \\\\ S=40*81\\\\ S=3240\end{gathered}a) S= 1+ 2+3+ ....+ 80=S=2[80(80+1)]S=40∗81S=3240
\begin{gathered}b)~S=~2~+~4~+6~+~...+100\\\\ S=2(1+2+3+...+50)\\\\ S= 2\frac{[50(50+1)]}{2} \\\\ S=50*51\\\\ S=2550\end{gathered}b) S= 2 + 4 +6 + ...+100S=2(1+2+3+...+50)S=22[50(50+1)]S=50∗51S=2550
\begin{gathered}c)~S=~1+3+5+...+99\\\\ 1=2*0+1\\ 3=2*1+1\\ 5=2*2+1\\ .........\\ 99=2*49+1\\\\ S=(2*0+1)+(2*1+1)+(2*2+ 1)+...+(2*49+1)\\ S=2*0+2*1+2*2+...+2*49+1+1+1+...+1\\ S=0+2(1+2+3+...+49)+50\\ S=2(49*50):2+50\\ S=49*50+50\\ S=2450+50\\ S=2500\end{gathered}c) S= 1+3+5+...+991=2∗0+13=2∗1+15=2∗2+1.........99=2∗49+1S=(2∗0+1)+(2∗1+1)+(2∗2+1)+...+(2∗49+1)S=2∗0+2∗1+2∗2+...+2∗49+1+1+1+...+1S=0+2(1+2+3+...+49)+50S=2(49∗50):2+50S=49∗50+50S=2450+50S=2500
Răspuns:
1+2+3+...+80=(1+80)*80 totul supra 2=81*80 totul supra 2(se simplifică 80 cu 2)=81*40=3240
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