# Thread: Is this sum property right?

1. ## Is this sum property right?

$\displaystyle \sum_{i=1}^{n} (n + i)^2 = (\sum_{i=1}^{n - 1} (n - 1 + i)^2) + (2n - 1)^2 + (2n)^2 - n^2$

Proof:

$\displaystyle \sum_{i=1}^{n} (n + i)^2 = (n + 1)^2 + (n + 2)^2 + \ldots + (2n - 3)^2 + (2n - 2)^2 + (2n - 1)^2 + (2n)^2 = n^2 + (n + 1)^2 + (n + 2)^2 + \ldots + (2n - 3)^2 + (2n - 2)^2 + (2n - 1)^2 + (2n)^2 - n^2$
$\displaystyle \sum_{i=1}^{n - 1} (n - 1 + i)^2 = n^2 + (n + 1)^2 + \ldots + (2n - 3)^2 + (2n - 2)^2$

So, $\displaystyle \sum_{i=1}^{n} (n + i)^2 = \sum_{i=1}^{n - 1} (n - 1 + i)^2 + (2n - 1)^2 + (2n)^2 - n^2$

2. ## Re: Is this sum property right?

The answer you gave appears correct. Expanded, the result can be written as:
$\displaystyle \sum_{i=1}^{n} (n + i)^2 = (\sum_{i=1}^{n - 1} (n - 1 + i)^2) + (2n - 1)^2 + (2n)^2 - n^2 = (\sum_{i=1}^{n - 1} (n - 1 + i)^2) + 7n^2 - 4n + 1$

My way
$\displaystyle \sum_{i=1}^{n} (n + i)^2 = \sum_{i=1}^{n-1} (n + i)^2 + (2n)^2$ (Take out the last term)
$\displaystyle = [\sum_{i=1}^{n-1} (n - 1 + i)^2 + 2n-1 + 2i] + (2n)^2$ (Expanded each square and compared)
$\displaystyle = \sum_{i=1}^{n-1} (n - 1 + i)^2 + 2 \sum_{i=1}^{n-1} i + (n-1)(2n-1) + (2n)^2$ (Removing (2n-1) from the sum means multiplying it n-1 times)
$\displaystyle = \sum_{i=1}^{n-1} (n - 1 + i)^2 + (n-1)n + (n-1)(2n-1) + (2n)^2$

The part outside the first sum is equal to what you have, which is
$\displaystyle 7n^2 - 4n + 1$

double post