summation notation: A grocer builds a mountain so that each

[btrfly24]

A grocer builds a mountain so that each level after the first contains one less row and one less can in each row. The first level has 9 rows and 12 cans in each row. Write a sequence whose terms are the number of cans at each level. Write the sum of the terms of this sequence in summation notation and find the number of cans in the mountain.

I've figured the number of cans as 9/2(108+4) = 504 cans. I'm hoping that this much is right, but I'm kinda stuck in writing the sum of the terms in summation notation.

+------+----+----+----+---+---+---+---+---+---+
| cans |  9 |  8 |  7 | 6 | 5 | 4 | 3 | 2 | 1 |
+------+----+----+----+---+---+---+---+---+---+
| rows | 12 | 11 | 10 | 9 | 8 | 7 | 6 | 5 | 4 |
+------+----+----+----+---+---+---+---+---+---+

Am I somewhat close or can someone help!!? Thanks in advance.

 

[soroban]

Re: summation notation

Hello, btrfly24!

The grocer builds a mountain so that each level after the first
contains one less row and one less can in each row.
The first level has 9 rows and 12 cans in each row.
(a) Write a sequence whose terms are the number of cans at each level.
(b)Write the sum of the terms of this sequence in summation notation.
(c) Find the number of cans in the mountain.

Now I've figured the number of cans as 9/2(108+4)=504 cans
Sorry, no . . . It is not an arithmetic series

\(\displaystyle \L\begin{array}{cccccccccc}\text{cans} & 9 & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\ \text{rows} & 12 & 11 & 10 & 9 & 8 & 7 & 6 & 5 & 4 \end{array}\)

Starting at the top, the number of cans at each level is:
. . \(\displaystyle \L1\cdot4,\:2\cdot5,\;3\cdot6,\;4\cdot7,\;5\cdot8,\;6\cdot9,\;7\cdot10,\,8\cdot11,\;9\cdot12\;\) [1]


(a) The sequence is: \(\displaystyle \L\:4,\;10,\;18,\;28,\;40,\;54,\;70,\;88,\;108\;\) [2]
. . .The differences are not constant.
. . .It is not an arithmetic progression.


(b) Look at [1].
. . .Each term is: .(a number) x (the number plus 3)
. . .So each term is of the form: .\(\displaystyle n(n\,+\,3)\) ... and \(\displaystyle n\) goes from \(\displaystyle 1\) to \(\displaystyle 9\).

. . .The total would be: \(\displaystyle \L\;\sum^9_1n(n\,+\,3)\)


(c) There is a way to construct a general formula the total number of cans,
. . .but it is faster to simply add the numbers in [2].

. . .\(\displaystyle \L4\,+\,10\,+\,18\,+\,28\,+\,40\,+\,54\,+\,70\,+\,88\,+\,108\;=\;\fbox{420}\)

 

[btrfly24]

OOOHH.......I guess if I could have just figured out the sequence of numbers I'd have been on the right tract. Thanks so much for explaining that!