Question

ajutatima va rog. e urgent​

Answer 1

a) 2+3+4+...+50= aplicam suma lui Gauss

$\frac{50(50 + 1)}{2} = \frac{50 \times 51}{2} = 25 \times 51 = 1275$

1275-1=1274

b) 1+2+3+....+500= aplicam suma lui gauss

$\frac{500(500 + 1)}{2} = \frac{500 \times 501}{2} = 250 \times 501 = 125250$

c) 1+97+2+999+3+198=

(1+999)+(97+3)+(198+2)=

1000+100+200=1300

d)1+2+3+...+124= aplicam suma lui gauss

$\frac{124(124 +1)}{2} = \frac{124 \times 125}{2} = 62 \times 125 = 7750$

e) 2+4+6+...+600=

2(1+2+3+....+300) )= aplicam suma lui gauss pentru: 1+2+3+....+300=

$\frac{300(300 + 1)}{2} = \frac{300 \times 301}{2} = 150 \times 301 = 45150$

2×45150= 90300

f)15+16+17+...+70=

(15+16+17+...+70) - (1+2+3+...+14)

aplicam suma lui gauss

•15+16+17+....+70=

$\frac{70(70 + 1)}{2} = \frac{70 \times 71}{2} = 35 \times 71 = 2485$

•1+2+3+...+14=

$\frac{14(14 + 1)}{2} = \frac{14 \times 15}{2} = 7 \times 15 = 105$

2485-105=2380

g)5+6+7+...+81=

(5+6+7+...+81) - (1+2+3+4)

aplicam suma lui gauss

•5+6+7+...+81=

$\frac{81(81 + 2)}{2} = \frac{81 \times 82}{2} = 81 \times 41 = 3321$

•1+2+3+4=10

3322-10=3312

h)1+2+3+...+512= aplicam suma lui gauss

$\frac{512(512 + 1)}{2} = \frac{512 \times 513}{2} = 256 \times 513 = 131328$

i)317+318+319+...+400=

(317+318+319+...+400) - (1+2+3+...+316)

aplicam suma lui gauss

$\frac{400(400 + 1)}{2} - \frac{316(316 + 1)}{2} = \frac{400 \times 401}{2} - \frac{316 + 317}{2} \\ = 200 \times 401 - 158 \times 317 = \\ 80200 - 50086 = 30114$