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Sagot :
Given:
The sum of the first 100 counting numbers is 5050.
To find:
The difference between the sum of all of the even counting numbers and the odd counting numbers less than 101.
Explanation:
Let us find the sum of all of the even counting numbers from 1 to 101.
The series is,
[tex]S_1=2+4+6+....+100[/tex]It can be written as,
[tex]S_1=2(1+2+3+.....+50)[/tex]Using the formula,
[tex]\begin{gathered} 1+2+3+.....+n=\frac{n(n+1)}{2} \\ S_1=2(1+2+3+....+50)=2[\frac{50(50+1)}{2}] \\ S_1=50(51) \\ S_1=2550........(1) \end{gathered}[/tex]Next, let us find the sum of all of the odd counting numbers.
[tex]\begin{gathered} S_2=Total-Sum\text{ of all even numebrs} \\ S_2=5050-2550 \\ S_2=2500.......(2) \end{gathered}[/tex]So, the difference between the sum of all of the even counting numbers and the odd counting numbers less than 101 is
[tex]\begin{gathered} S_1-S_2=2550-2500 \\ =50 \end{gathered}[/tex]Final answer:
The difference between the sum of all of the even counting numbers and the odd counting numbers less than 101 is 50.
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