Question

Find the sum of the first n even natural numbers. Hence find the sum of the first 20 even natural numbers.

Answer 1

The first n even numbers are 2, 4, 6, 8,…
Now, we observe that the numbers are in an A.P.
Here,
First term, a = 2
Common difference, d = t₂ – t₁ = 4 – 2 = 2
We know that the sum of n terms of an A.P. is $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$.
Thus, we get the sum of the first n even numbers as:
$$\begin{gathered} S_n=\frac{n}{2}\left[2\left(2\right)+(n-1)\left(2\right)\right] \\ =\frac{n}{2}\left[4+2n-2\right] \\ =\frac{n}{2}\times (2n+2) \\ =\frac{n}{2}\times 2(n+1) \\ =n(n+1) \end{gathered}$$

$$\begin{gathered} \mathrm{Now}, \\ \mathrm{To}\mathrm{find}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{first}20\mathrm{even}\mathrm{natural}\mathrm{numbers},\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{take}n=20. \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ {S}_{20}=20(20+1)=20\times 21=420 \end{gathered}$$