Prove by method of induction, for all $n \in N$: $2 + 4 + 6 + \ldots + 2n = n(n+1)$.

Prove by method of induction, for all n ∈ N: 2 + 4 + 6 + … + 2n = n(n+1).

EXERCISE 4.1 · Q1

n \in N

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Let $P(n):2+4+\cdots+2n=n(n+1)$ . Base case ( $n=1$ ): L.H.S. $=2$ , R.H.S. $=1(2)=2$ ; equal, so $P(1)$ holds. Inductive hypothesis: assume $P(k)$ : $2+4+\cdots+2k=k(k+1)$ . Inductive step: add $2(k+1)$ to both sides: $2+4+\cdots+2k+2(k+1)=k(k+1)+2(k+1)=(k+1)(k+2)$ , which is exactly $P(k+1)$ . Conclusion: by the Principle of Mathematical Induction, $P(n)$ holds for every $n\in N$ .

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2+4+6+\cdots+2n=n(n+1)

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Q.Prove by method of induction, for all $n \in N$ : $2 + 4 + 6 + \ldots + 2n = n(n+1)$ .

Concept understanding — Principle of Mathematical Induction

Q.Prove by method of induction, for all n∈Nn \in Nn∈N: 2+4+6+…+2n=n(n+1)2 + 4 + 6 + \ldots + 2n = n(n+1)2+4+6+…+2n=n(n+1).

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Q.Prove by method of induction, for all $n \in N$ : $2 + 4 + 6 + \ldots + 2n = n(n+1)$ .

Unlock everything free for 14 days