Question

Prove by the method of induction, for all $n \in \mathbb{N}$ $2 + 4 + 6 + .... + 2n = n(n + 1)$ .

Answer 1

For proving this by the method of induction at first, we have to show the result is true for $n = 1$ . Then we will suppose that the result is true for $n = k$ where $k$ is some natural number. Then we will prove the result is true for $n = k + 1$ . This will imply that the result is true for each and every natural number.
Used Principle:
Principle of induction:
Let $S$ be a subset of $\mathbb{N}$ with the properties –
$1$ belongs to $S$ , and
Whenever a natural number $k$ belongs to $S$ , then $k + 1$ belongs to $S$ .

Complete step-by-step solution:
For $n = 1$ , the statement is true because $2 = 1 \cdot (1 + 1)$ .
Let us assume that the statement is true for some natural number $k$ .
Then $2 + 4 + 6 + ..... + 2k = k(k + 1)$ .
Now for $n = k + 1$ we will get;
$2 + 4 + 6 + ..... + 2k + 2(k + 1)$
Now we already assumed that $2 + 4 + 6 + ..... + 2k = k(k + 1)$ .
Hence;
$2 + 4 + 6 + ..... + 2k + 2(k + 1) = k(k + 1) + 2(k + 1)$
Taking $k + 1$ common from the right-hand side we get;
$\Rightarrow 2 + 4 + 6 + .... + 2k + 2(k + 1) = (k + 1)(k + 2)$
In the statement, if we put $k + 1$ for $n$ , we will get the above result.
Hence the statement is satisfied for $n = k + 1$ .
By the principle of induction, the statement is true for all natural numbers $n$ .

Additional Information: The set of natural numbers is infinite. To define any number we use $n =$ some natural number. In the principle of induction we stop after showing the statement $n = k + 1$ because after that similarly, the statement will be true for $k + 2$ , $k + 3$ ,….. and so on.

Note: In the statement to denote every natural number, we use $n$ . But whenever we are taking $n = k$ the natural number $k$ is fixed. After that, we prove the result for $k + 1$ . Students should be careful about the case $n = k$ and $n = k + 1$ . The first case is assumed by us and the second case we have to prove by using the result for $n = k$ .