Question

Which of the following is equivalent to...

Answer 1

`\displaystyle \boxed{\sum_(n=1)^(19) (n^2 - 6n + 9) - \sum_(n=1)^(9) (n^2 - 6n + 9) = \sum_(n=10)^(19) (n-3)^2}.`

Explanation:

We can start by noticing that

`n^2 - 6n + 9 = n^2 - 2 * 3 * n + 3^2 = (n-3)^2,`

so we get:

`\displaystyle \sum_(n=1)^(19) (n^2 - 6n + 9) - \sum_(n=1)^(9) (n^2 - 6n + 9) = \sum_(n=1)^(19) (n-3)^2 - \sum_(n=1)^(9) (n-3)^2.`

Let
`a_n=(n-3)^2`. We can now write the expression as:

`\displaystyle\sum_(n=1)^(19) a_n - \sum_(n=1)^(9) a_n = (a_1 + a_2 + \dots + a_(19)) - (a_1 + a_2 + \dots + a_9).`

Since the first 9 terms cancel, we get:

`\displaystyle a_(10) + a_(11) + \dots + a_(19) = \sum_(n=10)^(19)a_n.`

So we finally get:

`\displaystyle \boxed{\sum_(n=1)^(19) (n^2 - 6n + 9) - \sum_(n=1)^(9) (n^2 - 6n + 9) = \sum_(n=10)^(19) (n-3)^2}.`