How is the base case $P(1)$ verified for the conjecture $2 + 4$ + ... + $2n = n(n+1)$ ?

Q: How is the base case P(1) verified for the conjecture 2 + 4 + ... + 2n = n(n+1)?

To verify the base case, substitute n=1 into both sides of the equation. The left-hand side (LHS) becomes the first term, 2. The right-hand side (RHS) becomes 1(1+1), which simplifies to 2. Since LHS equals RHS, P(1) is true.

To verify the base case, substitute

n=1

into both sides of the equation. The left-hand side (LHS) becomes the first term, 2. The right-hand side (RHS) becomes

1(1+1)

, which simplifies to 2. Since LHS equals RHS,

P(1)

is true.

Conditions

The proposition is $P(n)$ : $2 + 4$ + ... + $2n = n(n+1)$ .
The base case is checked at $n=1$ .

Reasoning, step by step

Identify the LHS for $n=1$ : The sum contains only the first even integer, so LHS = 2.
Identify the RHS for $n=1$ : Substitute $n=1$ into $n(n+1)$ to get $1(1+1)$ .
Simplify the RHS: $1(1+1) = 1(2) = 2$ .
Compare LHS and RHS: $2 = 2$ .
Conclude that $P(1)$ is true.

Example

The video shows: ' $P(1)$ _LHS = 2' and ' $P(1)$ _RHS = $1(1+1) = 2$ '. The narrator states, 'Since both sides equal 2, the base case is verified.'

Common misconceptions

Calculating the LHS as the sum of all terms up to 2n without setting $n=1$ .
Forgetting to simplify the RHS correctly.

Watch the explanation

YouTubePrinciple Of Mathematical Induction | Don't Memorise

3:42 – 4:19Watch this moment ↗

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Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.

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