Proof by induction

Question

This is a topic I always struggle with. Could someone help me prove this:

Prove by induction: $$ \sum_{k=1}^n 2k = n^2 + n, $$ Thanks for any help

Answer 1

Check for $n=1$. It is true.

Assuming it is true for $a$,

$$\begin{align} \sum_{k=1}^a 2k &= a^2 + a\\ \sum_{k=1}^a 2k+2(a+1) &= a^2 + a+2(a+1)\\ \sum_{k=1}^{a+1} &= a^2 + a+2(a+1)\\ &=(a^2+2a+1)+a+1\\ &=(a+1)^2+(a+1) \end{align}$$

It then follows that it is true for $a+1$. Then, because it is true for $1$, it is true for $1+1=2$, $2+1=3$, and so on. It is true for all positive integers.

Answer 2

Consider

Step 1: $n=1$ and try to show that your equation is true

Step 2: Let consider your equation is true for $n=m$ and try to show it is also true for $n= m+1$. Then you will be done.

Note: $\sum_{k=1}^n 2k = 2\times \frac{n^2 + n}2 = n^2 + n$. So without induction you can prove this easily.

Answer 3

Note that Base case holds easily.For next Step Hint $$ \sum_{k=1}^{n} 2k = \sum_{k=1}^{n-1} 2k +2n$$