tli8322 tli8322
  • 04-08-2022
  • Mathematics
contestada

How do I solve “1+2+3+4+5+…100=“
A. 1010
B. 5050
C. 5000
D. 1000

Respuesta :

jetsungyembo
jetsungyembo jetsungyembo
  • 04-08-2022

Answer:

5050

Step-by-step explanation:

we know that 100/2 is 50 and 50 x 100 is 5000

so now another 50 is remaining but we can't multiply but add so

1+2+3+4+5...100 = in simple form= 50x100+50=5050//

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