Question

What is the sum of first \[n\] even natural numbers?

Answer 1

Hint: Even numbers is a series of numbers with a difference between the adjacent numbers as \[2\] and also must be divisible by \[2\]. Means even numbers is a series of sequences of numbers which are divisible by \[2\] and have common difference \[2\]. To calculate the sum of these numbers we use the formula of sum of arithmetic progression for which the terms depend on \[n\] and hence sum also.

Formula used:
\[({{a}_{n}})={{a}_{1}}+(n-1)d\] and \[{{S}_{n}}=\dfrac{n}{2}\{2a+(n-1)d\}\]
Where, \[{{a}_{1}}\] is first term of series
\[n\] is number of terms
\[d\] is common difference
\[{{S}_{n}}\] is sum of \[n\] terms

Complete step by step solution:
Given that,
Total even numbers in sequence are \[n\] with first term is first even number and that is \[2\]
The series is \[2,4,6,8,10,{{...}_{n}}\]
\[\Rightarrow {{a}_{1}}=2\]
Now using the formula for sum of this series
\[{{S}_{n}}=\dfrac{n}{2}\{2a+(n-1)d\}\]
Now substituting the values
\[{{S}_{n}}=\dfrac{n}{2}\{2\times 2+(n-1)2\}\]
(Since the total terms are \[n\] as we have to calculate the sum of first \[n\] even numbers)
\[\Rightarrow {{S}_{n}}=\dfrac{n}{2}\{4+(n-1)2\}\]
\[\Rightarrow {{S}_{n}}=\dfrac{n}{2}\{4+2n-2\}\]
\[\Rightarrow {{S}_{n}}=\dfrac{n}{2}\{2+2n\}\]
Taking common \[2\] from bracket
\[\Rightarrow {{S}_{n}}=n(n+1)\]

Hence, we have calculated the sum of first \[n\] even numbers and that is \[n(n+1)\]

Note:
First find a general term \[({{a}_{n}})\] of this sequence and this would be in terms of n and using this we will calculate the last term and we have a formula to calculate the sum of the series of arithmetic progression when the first and last term and number of terms given and that is \[S=\dfrac{n}{2}({{a}_{1}}+{{a}_{n}})\]
Here, Sum is the product of the average of first and last term with the total number of terms.