Arătați că:
a)1+3+5+7+9+11=
b)1+3+5+7+...+19=
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Răspunsuri la întrebare
a) 1 + 3 + 5 + 7 + 9 + 11 =
= ( 1 + 11 ) + ( 3 + 9 ) + ( 5 + 7 ) =
= 12 + 12 + 12 =
= 36
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1 + 3 + 5 + 7 + 9 + 11 = 6 × ( 1 + 11 ) : 2 = 6 × 12 : 2 = 72 : 2 = 36 → aplicand formula sumei lui Gauss
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b) 1 + 3 + 5 + 7 + ........ + 19 =
= ( 1 + 19 ) + ( 3 + 17 ) + ( 5 + 15 ) + ( 7 + 13 ) + ( 9 + 11 ) → aplic asociativitatea
= 20 + 20 + 20 + 20 + 20 = → 5 ori cate 20
= 100
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S = 1 + 3 + 5 + 7 + ...... + 19
→ aflu cati termeni are suma numerelor consecutive impare ( ratia = 2)
3 > 1 cu 2; 5 - 3 = 2 .....
( 19 - 1 ) : 2 + 1 = 18 : 2 + 1 = 9 + 1 = 10 termeni
→ aplic formula sumei lui Gauss
S = 10 × ( 1 + 19 ) : 2
S = 200 : 2
S = 100
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sau:
S = 1 + 3 + 5 + 7 + ....... + 19 → suma primelor 10 numere impare consecutive
= 1 + ( 1 + 2 ) + ( 1 + 4 ) + ( 1 + 6 ) + ...... + ( 1 + 16 ) + ( 1 + 18 ) =
= 1 × 10 + 2 × ( 1 + 2 + 3 + ..... + 9 ) =
= 10 + 2 × 9 × 10 : 2 =
= 10 + 90 =
= 100