Question
Write the sum of first $n$ even natural numbers.
✓
Answer
Let,
Even numbers are, $2, 4, 6, 8, .....$
Here,
First term $a = 2$
Difference $d = 4 - 2 = 2$
We know. Sum of $n$ terms,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}[2(2)+(\text{n}-1)2]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}[4+2\text{n}-2]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}[2\text{n}+2]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}\times2[\text{n}+1]$
$\Rightarrow\ \text{S}_\text{n}=\text{n}(\text{n}+1)$
Hence, Sum of even numbers is $n(n + 1).$
Even numbers are, $2, 4, 6, 8, .....$
Here,
First term $a = 2$
Difference $d = 4 - 2 = 2$
We know. Sum of $n$ terms,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}[2(2)+(\text{n}-1)2]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}[4+2\text{n}-2]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}[2\text{n}+2]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}\times2[\text{n}+1]$
$\Rightarrow\ \text{S}_\text{n}=\text{n}(\text{n}+1)$
Hence, Sum of even numbers is $n(n + 1).$
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