2
$\begingroup$

The following is a question which has been bugging me for quite a while,

Find the sum of the series $$1^2 + 2 × 2^2 + 3^2 + 2 × 4^2 + 5^2 + 2 × 6^2 + . . . + 2(n − 1)^2 + n^2 ,$$ Where $n$ is odd

I started by denoting this entire series with $S$, from there it is apparent that $S$ infact consists of two lesser series I shall call;

  1. $S_A$ The sum of the squares of odd natural numbers up till $n^2$
  2. $S_B$ Twice the sum of the squares of even natural numbers up till $(n-1)^2$

I resolved this would be easier to tackle by noting that $$S_A + \frac{1}{2}S_B = \sum_{r=1}^{n} r^2$$ Which is just the sum of the squares of the first $n$ natural numbers $$\therefore S_A +\frac{1}{2}S_B=\frac{n}{6}(n+1)(2n+1)$$

Leaving only the value of $\frac{1}{2}S_B$ to be found, this is where I am currently facing difficulty as I am unsure on whether my working is correct;

For the sum of the squares of the first n even natural numbers;

$$2^2 + 4^2 .... (2n)^2=2^2\sum_{r=1}^{n} r^2$$

$$\implies \frac{2}{3}n(n+1)(2n+1) $$

Hence the sum of the first $n-1$ even natural numbers should be

$$ \frac{2}{3}n(n-1)(2n-1)$$

And

$$S= \frac{n}{6}(n+1)(2n+1) + \frac{2}{3}n(n-1)(2n-1)$$

$$\therefore S= \frac{1}{6}n(10n^2 -9n + 5n) $$ However the correct answer is

$$\frac{1}{2}n^2(n+1)$$

Where has my working gone wrong and how would I arrive at the correct answer?

$\endgroup$

3 Answers 3

2
$\begingroup$

The given sum is equal to $$\begin{align} 1^2 + 2 × 2^2 + 3^2 + &2 × 4^2 + 5^2 + 2 × 6^2 + \dots + 2(n − 1)^2 + n^2\\ &=1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + \dots + (n − 1)^2 + n^2\\ &\qquad\;+2^2 + 4^2 + 6^2 + \dots + (n − 1)^2\\ &=\sum_{k=1}^n k^2+\sum_{k=1}^{m}(2k)^2=\sum_{k=1}^n k^2+4\sum_{k=1}^{m}k^2\\ &=\frac{n(n+1)(2n+1)}{6}+4\frac{m(m+1)(2m+1)}{6}\\&=\frac{n(n+1)(2n+1)}{6}+\frac{(n-1)(n+1)n}{6}\\ &=\frac{n(n+1)(2n+1+n-1)}{6}=\frac{n^2(n+1)}{2} \end{align}$$ where $m=\frac{n-1}{2}$. Note that in the sum involving the even squares there are $m$ terms (not $(n-1)$).

$\endgroup$
2
  • $\begingroup$ Can you explain why the second partial summation is up to $m=\frac{n-1}{2}$ and not $n-1$? $\endgroup$ Aug 16, 2021 at 16:42
  • $\begingroup$ @Filthyscrub The last term in the second sum is $(n-1)^2=(2m)^2$ $\endgroup$
    – Robert Z
    Aug 16, 2021 at 16:45
2
$\begingroup$

As implicitely pointed out in the given answer, your mistake is due to the fact that for the sum of the squares of even natural numbers up to $n-1$ we need to consider the following:

$$2^2 + 4^2 .... (n-1)^2=2^2\left(1+2^2+\dots+\left(\frac{n-1}2\right)^2\right) $$

and not

$$2^2 + 4^2 .... (2n)^2=2^2\sum_{r=1}^{n} r^2$$

which is the sum of the squares of even natural numbers up to $2n$.

$\endgroup$
1
$\begingroup$

Let f(n) be the sum of squares from 1 up to $n^2$. Then the answer to your question is f(n) + 4 f((n-1)/2). Find a formula for f(n) and substitute it. It should be close to 1.5 f(n) since you are adding about half of the terms twice.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .