Răspuns :

Targoviste44id

[tex]\it 1+2+3+\ ...\ +n=\dfrac{n(n+1)}{2}\ \ Formula\ \ lui\ \ Gauss\\ \\ \\ a)\ \ 1+2+3+\ ...\ +54=\dfrac{54\cdot55}{2}=27\cdot55=1485\\ \\ \\ b)\ \ 1+2+3+\ ...\ +13=\dfrac{13\cdot14}{2}=13\cdot7=91\\ \\ \\ c)\ \ 1+2+3+\ ...\ +200=\dfrac{200\cdot201}{2}=100\cdot201=20100\\ \\ \\ d)\ \ 1+2+3+\ ...\ +97=\dfrac{97\cdot98}{2}=97\cdot49=4753\\ \\ \\ e)\ \ 4+5+6+\ ...\ +200=1+2+3+\ ...\ +200-(1+2+3)=\dfrac{200\cdot201}{2}-6=\\ \\ \\ =20100-6=20094[/tex]

[tex]\it f)\ \ 2+4+6+\ ...\ +42=2\cdot(1+2+3+\ ...\ +21)=\not2\cdot\dfrac{21\cdot22}{\not2}=21\cdot22=[/tex]

[tex]\it g)\ \ 3+6+9+\ ...\ +99=3\cdot(1+2+3+\ ...\ +33)=3\cdot\dfrac{33\cdot34}{2}=3\cdot33\cdot17=[/tex]

Răspuns:

Explicație pas cu pas:

Suma lui Gauss este 1+2+3+...+n= n(n+1)/2

1) 1+2+3+4+....+54=54·55/2=27·55=1485

2) 1+2+3+...+13= 13·14/2=13·7=91

3) 4+5+6+...+200= (1+2+3+4+5+6+...+200)-(1+2+3)=200·201/2-3·4/2=20100-6= 20094

4)11+12+13+....97 = (1+2+3+...+97)-(1+2+....10)=97·98/2 - 10·11/2 =4753-55 = 4698

5) 32+33+34+...+112= (1+2+3+....+112) - (1+2+3+...+31) = 112·113/2 - 31·32/2 = 6272 - 496 =5776

6) 1+2+3+...+876 = 876·877/2=384126

7) 925+926+927+...+1032 = (1+2+3+...+1032)- (1+2+3+....924) = 1032·1033/2 - 924·925/2 =533028 - 427350 =105678

8) 72+73+74+...+204= (1+2+3+...+204)- (1+2+3+...+71) = 204·205/2 - 71·72/2 =

20910-2556 =18354

9) 2+4+6+...+42 = 2·(1+2+3+....+21) =2·21·22/2=21·22=462

10) 3+6+9+....+99= 3·(1+2+3+....+33) = 3·33·34/2 =1683

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