Question

What is the sum of first $n$ even natural numbers?

Answer 1

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.