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Blondaviperal
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Sa se calculeze suma 1+3+5+...+21
1+11+21+31+...+111 sunt progresii clasa 9

Răspuns :

Ioanna
a.)1 + 3 + 5 + ... + 21
a₁=1    a₂=3    ⇒  r = 3-1 = 2
Sn = [tex] \frac{(a1 + an)n}{2} [/tex]n
S₁₁ =  [tex] \frac{(a1 + a11)11}{2} [/tex] =  [tex] \frac{(1 + 21)11}{2} [/tex] = [tex] \frac{22ori11}{2} [/tex] =  11 · 11 = 121
   an = a₁ + (n - 1)r    ⇒ a₁₁ = 1 + (11-1)2 = 1 + 10 · 2 = 21
b.)1 + 11 + 21 + ... + 111
a₁=1    a₂=11    ⇒  r = 11-1 = 10
S₁₂ =  [tex] \frac{(a1 + a12)12}{2} [/tex] =  [tex] \frac{(1 + 111)12}{2} [/tex] = 112 · 6 = 672
     a₁₂ = 1 + (12-1)10 = 1 + 11 · 10 = 111



1+ 3+ 5+ 7+ 9+ ...+ 19+ 21=
1+(2+1)+ (4+1)+ ...+(18+1)+(20+ 1)=
(1+ 1+ 1+...+1)+ ( 2+ 4+ 6+...+18+ 20)=
(1+ 1+ 1+...+1)+ ( 2·1+2·2 + 2·3+...+2·9+ 2·10)=
1·11+2·(1+ 2+ 3+  +9+ 10)=
11+ 2·10·11:2=     Se reduc!
11+10·11=
11·1+10·11=
11(1+10)=
11·11=
121

1+11+21+31+...+111=
1+ (10+ 1)+ (20+ 1)+ (30+ 1)+ ...+(110+ 1)=
(1+ 1+ 1+   +1) +(10+ 20+ 30+ ...+ 110)=
(1+ 1+ 1+   +1) +10(1+ 2+ 3+ ...+ 11)=
1·12+10·11·12:2=
1·12+11·12·5=
12(1+11·5)=
12(1+55)=
12·56=
672

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