Question

Multiple Choice Question.
Show Work Needed.

Answer 1

First of all, every number in each row is odd, so we can eliminate 392 and 394.

The number of elements in each row forms an arithmetic sequence:

• 1st row : 1 element

• 2nd row : 3 elements - and 3 - 1 = 2

• 3rd row : 5 elements - and 5 - 3 = 2

• 4th row : 7 elements - and 7 - 5 = 2

and so on, so that the
`n`-th row has
`1 + 2(n-1) = 2n - 1` elements.

This means that in
`n-1` complete rows, there is a total of

`\displaystyle \sum_(i=1)^(n-1) (2i-1) = n^2 - 2n + 1`

elements, so that the first element in the subsequent
`n`-th row is the
`(n^2-2n+2)`-th number in the sequence {1, 3, 5, 7, …}, i.e the
`(n^2-2n+2)`-th odd positive integer. To compute the sum, I use the following well-known formulas.

`\displaystyle \sum_(i=1)^n 1 = n`

`\displaystyle \sum_(i=1)^n i = \frac{n(n+1)}2`

Now, the
`n`-th term of the sequence {1, 3, 5, 7, …} is simply
`2n-1`, the
`n`-th odd positive integer. So the first element in the
`n`-th row is

`2(n^2-2n+2) - 1 = 2n^2 - 4n + 3`

and hence the 1st element of the 15th row is

`2\cdot15^2 - 4\cdot15 + 3 = \boxed{393}`