BANKING Content

IB Security 2019 Question 5

Question 5

Sum of all positive integers from 1 to 100 is:

Solution

Sum of numbers from 1 to 100 $$=\dfrac{n(n+1)}{2}$$ where n = number of terms from 1 to 100 = 100.
Therefore, Sum of numbers from 1 to 100 $$=\dfrac{100\times101}{2} = 5050$$



Create a FREE account and get:

  • Download Maths Shortcuts PDF
  • Get 300+ previous papers with solutions PDF
  • 500+ Online Tests for Free

Comments
cracku

Boost your Prep!

Download App