Question

Find the sum of even positive integers between 1 and 200.

Answer 1

Hint: We have to find the sum of the first 100 even integers. We know that every even integer has 2 as common multiple. So, in each and every term we will take 2 as common. We know that the sum of n natural numbers is equal to \[\dfrac{n(n+1)}{2}\]. By using this formula, we can sum the first 100 even integers.

Complete step-by-step answer:
Before solving the problem, we have to know that an even number is a number which is said to be a multiple of 2. Hence, 2,4,6,8,10,12,14,16,18,20, …. are said to be even numbers.
In the question, we are given to find the sum of even positive integers between 1 and 100.
So,
\[Sum=2+4+6+8+10+12+14+..........+196+198+200\]
We know that even numbers are multiples of 2. So, we can take 2 as common from each and every term of the sum.
\[Sum=2(1+2+3+4+5+...........+98+99+100)\]
We know that the sum of n natural numbers is equal to \[\dfrac{n(n+1)}{2}\].
\[\Rightarrow 1+2+3+4+5+......+n=\dfrac{n(n+1)}{2}\]
So, by using this formula
\[\begin{align}
  & Sum=2\left( \dfrac{100(100+1)}{2} \right) \\
 & \Rightarrow Sum=(100)(101) \\
 & \Rightarrow Sum=10100 \\
\end{align}\]
Therefore, the sum of 100 even positive integers is equal to 10100.

Note: This problem can be solved in an alternative method also.
We know that the sum of 100 even integers is equal to
\[Sum=2+4+6+8+10+12+14+..........+196+198+200.....(1)\]
In equation (1),
We are having the first term=2
In the similar way, we are having second term=4
In the similar manner, we are having third term =6
In the similar manner, we are having fourth term =8
In the similar manner, we are having \[{{n}^{th}}\]term =2n
We can observe that the
Difference between second term and first term = 4-2=2
In the similar manner, difference between third term and second term = 6-4=2
In the similar manner, difference between fourth term and second term =8-6=2
So, we can say that the common difference between each and every term is equal to 2.
So, we can say that the terms in the sum are in A.P.
We know that if a is first term, d is a common difference and n is number of terms, then the sum of terms of A.P is equal to \[\dfrac{n}{2}[2a+(n-1)d]\].
From this sum, we can say that n=100, a=2 and d=2.
So,
From equation (1), we get
\[\begin{align}
  & Sum=2+4+6+8+10+12+14+..........+196+198+200 \\
 & \Rightarrow Sum=\dfrac{100}{2}[2(2)+(100-1)2]=50[202]=10100 \\
\end{align}\]
So, we can say that the sum of the first 100 even integers is equal to 10100.