Prove even integer sum using induction

Question

This is a homework problem, so please do not give the answer away. I must prove the following using mathematical induction:

$\forall n\in\mathbb{Z^+},\;2+4+6+\cdots+2n=n^2+n.$

This is what I have so far:

Let $P\left(n\right)$ represent $2+4+6+\cdots+2n=n^2+n$. Since $P\left(1\right)=\left(1\right)^2+1=2$, $P\left(1\right)$ is true. If $P\left(k\right)$ is true for $k\in\mathbb{Z^+}$, then $P\left(k+1\right)$ is true. Hence \begin{align} n^2+n & = \left(k+1\right)^2+\left(k+1\right)\notag\\ & = k^2 + 2k + 1 + k + 1\notag\\ & = k^2+3k + 2\notag\\ &=\left(k+2\right)\left(k+1\right).\notag \end{align}

I feel the last few steps do not do justice. Could you provide a hint to get me back on track? Thank you for your time,

Answer 1

A little late to the party here, but I thought I might weigh in to share something useful I learned about summation/product identities when I first learned about induction, especially since this question has the proof-writing tag (I'm only going to focus on the inductive step, as you have already established the base case).

For most basic problems, the idea is to "peel off the $k+1$th summand" and then use the inductive hypothesis appropriately, etc. I'll give you an example concerning your problem specifically (since you have already accepted an answer, I imagine you have already proved the result).

You are trying to show that $$ 2+4+6+\cdots+2n=n^2+n\tag{1} $$ for all $n\geq 1$, where $n\in\mathbb{Z^+}$. Notice that we can represent $(1)$ by using $\Sigma$-notation: $$ \sum_{i=1}^n 2i=n^2+n.\tag{2} $$ Thus, we are really trying to prove that $(2)$ holds for all $n\in\mathbb{Z^+}$. To this end, let $P(n)$ denote the statement $$ P(n) : \sum_{i=1}^n 2i=n^2+n. $$ Fix some $k\geq 1$ and assume $P(k)$ to be true; that is, $$ P(k) : \sum_{i=1}^k 2i=k^2+k $$ holds. To be shown is that $P(k+1)$, where $$ P(k+1) : \sum_{i=1}^{k+1} 2i=(k+1)^2+(k+1), $$ follows. Beginning with the left-hand side of $P(k+1)$, \begin{align} \sum_{i=1}^{k+1} 2i &= \underbrace{\sum_{i=1}^k 2i + 2(k+1)}_{\text{"peel off the $k+1$th summand"}}\tag{by definition of $\Sigma$}\\[1em] &= (k^2+k) + 2(k+1)\tag{by $P(k)$, the ind. hyp.}\\[1em] &= (k+1)^2+(k+1),\tag{manipulate expression} \end{align} we end up with the right-hand side of $P(k+1)$.

Thus, by mathematical induction, the statement $P(n)$ holds for all $n\geq 1$. $\blacksquare$

Answer 2

Suppose $P(k)$ is true for some $k \in \mathbb{N}$. That is, \begin{equation} 2+4+6+\cdots+2k=k^2+k. \end{equation} Now, we need to use this to show \begin{equation} 2+4+6+\cdots+2k+2(k+1)=(k+1)^2+(k+1). \end{equation} I hope this helps.

Answer 3

I believe this might help you.

Hint: $\sum2n=2\sum n$, but $\sum n=\frac{n^{2}+n}{2}$.