# Mathematical Induction: Sum of first n even perfect squares [duplicate]

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So the series is $$P_k: 2^2 + 4^2 + 6^2 + ... + (2k)^2 = \frac{2k(k+1)(2k+1)}3$$

and i have to replace $P_k$ with $P_{k+1}$ to prove the series.

I have to show that $$\frac{2k(k+1)(2k+1)}3 + [2(k+1)]^2 = \frac{2(k+1)(k+2)(2k+3)}3$$ but I don't know how.

## marked as duplicate by MJD, qwr, Daniel Fischer, JimmyK4542, apnortonOct 6 '14 at 1:24

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• Note also that for the induction to go through, you will have to show that $P_1$ is true. I would expand the square, put everything over denominator $=3$ and take out a factor of $2(k+1)$. You should then have a quadratic expression to work with, which you should be able to factorise using a standard method. – Mark Bennet Oct 5 '14 at 19:01

## 3 Answers

$$\frac{2k(k+1)(2k+1)}{3}+(2(k+1))^2=\frac{2k(k+1)(2k+1)}{3}+\frac{3(2(k+1))^2}{3}=\\= \frac{2k(k+1)(2k+1)+3 \cdot 4(k+1)^2}{3}=\frac{2(k+1)(k(2k+1)+6(k+1))}{3}=\\=\frac{2(k+1)(2k^2+k+6k+6)}{3}=\frac{2(k+1)((2k+3)(k+2))}{3}$$

While I think you should be able to show that the identity holds using the usual algebraic manipulations (and if you have difficulties with that, you should practice this more), there are shortcuts that can save time and effort. If both sides are polynomials of the same nth degree (if they are not then they can't possibly be identical), then you only need to check that they agree with each other for n+1 values of the argument. In some cases it is easy to check that the coefficient of the largest power of the argument is the same on both sides and then you only need to consider n values of the argument.

In this case, you can very easily check that for k = 0, k = -1, and k = 1 both sides are the same and it's very easy to see that the coefficient of k^3 is the same. This takes less than a minute of mental effort and the simplicity of this method makes this almost guaranteed to be error proof.

There isn't a simple trick to solve this; it requires a battle with the algebra monster. When doing a proof by induction you have to start with the base case though, so let's do that first. We need to show that $$\sum_{i=1}^n(2i)^2=\frac{2n(n+1)(2n+1)}{3}$$

Base case n=1: $$\sum_{i=1}^1(2i)^2=4 = \frac{12}{3} = \frac{2 \cdot 2 \cdot 3}{3} = \frac{2\cdot 1(1+1)(2\cdot 1+1)}{3} \quad \checkmark$$

Induction Step: Suppose for all $k$ where $1 < k\leq n$ that $$\sum_{i=1}^k(2i)^2=\frac{2n(n+1)(2n+1)}{3}$$ Then we know $$\sum_{i=1}^{k+1}(2i)^2=\sum_{i=1}^{k}(2i)^2+ [2(k+1)]^2$$ Now we can apply the induction hypothesis to the above equality to make the step

$$\sum_{i=1}^{k}(2i)^2+ [2(k+1)]^2 = \frac{2k(k+1)(2k+1)}3 + [2(k+1)]^2 \\= \frac{2k(k+1)(2k+1)}3+ \frac{3[2(k+1)]^2}{3} \\ = \frac{2k(k+1)(2k+1)+3[2(k+1)]^2}3 \\ = \frac{2k(k+1)(2k+1)+12(k+1)^2}3$$ Factor out the quantity $2(k+1)$. $$= \frac{2(k+1)[k(2k+1)+6(k+1)]}{3} \\= \frac{2(k+1)(2k^2+k+6k+6)}{3} \\ = \frac{2(k+1)(2k^2+7k+6)}{3} \\= \frac{2(k+1)(k+2)(2k+3)}{3}\quad \quad \Box$$