Math Problem?

Question

It says "Consider the following table":

1 1 1
1 2 3 2 1
1 3 6 7 6 3 1
1 4 10 16 19 16 10 4 1

Find a pattern and write out the next 3 rows.

Can anyone help? Thanks.

Answer 1

1 1 1
1 2 3 2 1
1 3 6 7 6 3 1
1 4 10 16 19 16 10 4 1

1 5 15 30 45 51 45 30 15 5 1
1 6 21 50 90 126 141 126 90 50 21 6 1
1 7 28 77 161 266 357 393 357 266 161 77 28 7 1

Brief explantion:
each row starts and ends with a 1. The number of elements in each row increases by 2 each time. To find the second element of a row, add the first 2 elements of the row above it. To find the 3rd element of a row, add the the first 3 numbers of the row above it. Then each number after is the sum of three numbers in the previous row. Add from left to right. The pattern repeats at the midway point. I will explain the 5th row below...you need to look at the 4th row to see where I'm getting the numbers from:
1 4 10 16 19 16 10 4 1
1 5 15 30 45 51 45 30 15 5 1

so 5th row:
1 = 1
5 = 1+4
15 = 1+4+10
30 = 4+10+16
45 = 10+16+19
51 = 16+19+16 <--- midway point...
45 = 45
30 = 30
15 = 15
5 = 5
1 = 1

Answer 2

1 1 1
1 2 3 2 1
1 3 6 7 6 3 1
1 4 10 16 19 16 10 4 1
1 5 15 30 45 51 45 30 15 5 1
1 6.21.50.90.
1 7

I dont have the time to work it all out, but if you re postion the table like so (im assuming the middle row went 1, 3 7 19(yahoo doesnt even do a simple thing like formatting))add the the three numbers in the row above, eg
3rd line, i got 10 by adding 1, 3 and 6 from the row above. Under stand? Better do, i was thinking for a while...

Answer 3

1 5 15 30 45 51 45 30 15 5 1
1 6 21 50 90 126 141 126 90 50 21 6 1
1 7 28 77 161 266 357 393 357 266 161 77 28 7 1

Answer 4

1 4 10 16 19 16 10 4 1
1 5 15 30 45 51 45 30 15 5 1

"you should add the left 3 numbers of the previous row that are above the number that you are calculating"

i dont like it to do that for you its errorprone...
.

Answer 5

The next row is

1 5 15 30 45 49 45 30 15 5 1; the next one is

1 6 21 50 90 126 141 126 90 50 21 6 1; and the 3rd one is:

1 7 28 77 161 266 357 393 357 266 161 77 28 7 1.

Here is the rule, as well as how to find it:

You will notice that 111 * 111 = 12321, and that 111 * 12321 = 1367631. So multiplying by 111 generates the 2nd and 3rd rows. However, the next line shows that this particular rule cannot continue, as one will never get more than a single digit in any one column! But that very observation gives one the necessary clue:

Write down any row. Underneath it, write it out again, shifted right by one place. Below that, write it down once again, shifted over yet one more place.

Now: Add up, in COLUMNS, but WITHOUT "carrying" anything to another column. THAT gives you the fourth line,

1 4 10 16 19 16 10 4 1,

and continues to generate the next three lines I've given you.

Other "rules" about "adding up UP TO the three last numbers including the one in the column you're occupying" will work, but to my taste the fact that the first and last two columns do require special consideration, relative to the other columns, is an unaesthetic drawback. The idea of shifting the original row over by one place and two places, and then simply adding in columns without "carrying," seems much cleaner and more satisfying. [By reading the "rules" produced by the other responders, you may see how convoluted their descriptions can become!]

Live long and prosper.

Answer 6

Obviously start with 1, then add first two numbers of the previous row, then add first three. After that you just add threes. Thus the next row is

1 1+4 1+4+10 4+10+16 10+16+19 16+19+16 etc.
i.e.
1 5 15 30 45 51 45 30 15 5 1
Get it? So you can do the next two.

Answer 7

First net row:
1,5,15,30,45,51,45,30,15,5,1

Pattern:
a b C b a
a b c D c b a
a b c d E d c b a

Notes:
0) First digit is 1

1) Pattern is 2 digits longer each progressive row

2) The pattern MAXes at one more digit each row

3) The pattern is additive only up to the 3rd digit

4) The patern is both addative and subtractive from digit 3 to MAX digit.

5) The pattern just repeats backwards, recounting after the MAX digit.

So:
1,5,15,30,45,51,45,30,15,5,1

Was gotten by:
1,4,10,16,19,16,10,4,1
1 (1+4) (5+10) (15+16-1) (30+19-4) (45+16-10) (then repeat backwards)

Good luck.

Answer 8

Patterns are funny things.
Here's what I see.
First number always 1.
Second number sum of first 2 numbers from row above.
Third number sum of first 3 numbers from row above.
Fourth number sum of second thru fifth numbers from row above. (If there is no number it's zero)
Fifth number sum of third thru sixth numbers from row above
Sixth number sum of fourth thru seventh numbers from row above.

Carry on.
1, 5, 15, 30, 45, 51, 45, 30, 15, 5, 1
1, 6, 21,50, 90, 126, 141, 126, 90, 50, 21, 6, 1

Answer 9

These rows are the coefficients of the expansion of

(x^2+x+1) ^n for n=1,2,3,4,.......In order to see the pattern more easily,arrange the triangle so that the middle terms of
each row are right on top of each other. Then you can see that each term is the sum of the term just on top of it, and the two
terms on either side of that term. I leave it to others to write the rows.