Question

1.Calculati urmatoarele sume:
a)1+2+3+....+47+48
b)1+2+3+....+133+134
c)0+1+2+3+.....+94+95
Dau coroana!​

Answer 1

Răspuns:

Explicație pas cu pas:

1.Calculati urmatoarele sume:

Se foloseste formula sumei lui Gauss : 1+2+...+n=n(n+1)/2

a)1+2+3+....+47+48=48·49/2=24·49=1176

b)1+2+3+....+133+134=134·(134+1)/2=134·135/2=67·135=9045

c)0+1+2+3+.....+94+95=0+95(95+1)/2=0+95·48=0+4560=4560

Answer 2

Explicație pas cu pas:

sume Gauss:

d)

$1 + 2 + 3 + ... + 35 = \frac{35 \times (35 + 1)}{2} = \frac{35 \times 36}{2} = \bf 630$

e)

$1 + 2 + 3 + ... + 1000 + 1001 = \frac{1000 \times (1000 + 1)}{2} = \frac{1000 \times 1001}{2} = \bf 500500$

f)

$70 + 69 + ... + 3 + 2 + 1 = 1 + 2 + 3 + ... + 69 + 70 = \frac{70 \times (70 + 1)}{2} = \frac{70 \times 71}{2} = \bf 2485$

g)

$201 + 200 + ... + 3 + 2 + 1 = 1 + 2 + 3 + ... + 200 + 201 = \frac{201 \times (201 + 1)}{2} = \frac{201 \times 202}{2} = \bf 20301$

h)

$2 + 4 + 6 + ... + 58 + 60 = 2 \cdot (1 + 2 + 3 + ... + 28 + 30) = 2 \cdot \frac{30 \cdot 31}{2} = 30 \cdot 31 = \bf 930$