Question
Find the sum of first n even natural numbers.
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Answer
The first n even natural numbers are 2, 4, 6, 8, 10, ..., n.
Here, a = 2 and d = (4 - 2) = 2
Sum of n terms of an AP is given by
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$=\big(\frac{\text{n}}{2}\big)\times\big[2\times2+(\text{n}-1)\times2\big]$
$=\big(\frac{\text{n}}{2}\big)\times\big[4+2\text{n}-2\big]=\big(\frac{\text{n}}{2}\big)\times(2\text{n}+2)=\text{n}(\text{n}+1)$
Hence, the required is n(n + 1).
Here, a = 2 and d = (4 - 2) = 2
Sum of n terms of an AP is given by
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$=\big(\frac{\text{n}}{2}\big)\times\big[2\times2+(\text{n}-1)\times2\big]$
$=\big(\frac{\text{n}}{2}\big)\times\big[4+2\text{n}-2\big]=\big(\frac{\text{n}}{2}\big)\times(2\text{n}+2)=\text{n}(\text{n}+1)$
Hence, the required is n(n + 1).
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