What is the nth row in Pascal's Triangle?

Answer 1

As you may know, Pascal's Triangle is a triangle formed by values. The values increment in a predictable and calculatable fashion. Each value in a row is the sumb of the two values above it to the left and right. An example triangle to row 4 looks like:

0: 1

1: 1 1

2: 1 2 1

3: 1 3 3 1

4: 1 4 6 4 1

We will be using two variables: n for the row we will be working with, and k for the index of the value we are trying to find in any given row. It is important to note that we will be counting from 0 during this process (a common practice in computer science), so that what you might normally call the "first" row, we will actually be referring to as row 0 (n=0). Moreover, if we are evaluating for the sixth value in a row n, then the index is 6 and k=6 (although it is the seventh number in the row).

First of all, each row begins and ends with a 1 and is made up of (n+1) values.

For example, the "third" row, or row 2 where n=2 is comprised of "1 2 1". Since this is row 2, there should exist 2+1=3 values, the first and last of which are 1. Of course we can see that this is indeed true. Now let's find out why that middle number is 2.

In the special base cases of row 0 and row 1, the values are simply "1" in the former and "1 1" in the latter.

Method 1) After row 1, we need to use a formula to find values other than the 1's. This slightly-complex equation is (V_n,k)=(n!)/[k!(n-k)!]. This basically means that the spot represented in row n by index k is the value V. This number can be computed more easily than it might seem. (Now look at the bottom of this article for a general example. Keep reading to learn more than your fair share about Pascal's Triangle.)

Hint: The number after the first 1 and the number before the last 1 are both the same and are equal to n. This because n!/[1!(n-1)!] is equal to [n(n-1)!]/[(n-1)!] and simplifies to n once the (n-1)!'s cancel.

Ex1: What is the value of V?

0: 1

1: 1 1

2: 1 2 1

3: 1 3 3 1

4: 1 4 V 4 1

V_n,k = V_4,2 = n!/[1!(n-1)!] = 4!/[2!(4-2)!] = (4*3*2!)/(2!2!) = 12/2 = 6

Look above to see that we've performed the operations successfully.

ALTERNATE METHODS:

Method 2) Choosing

This is the simplest method of all, but only works well if you already have a calculator. To find the value V_n,k = V_7,4 plug n and k into the Choose operator. Written, this looks like (7c4), but some calculators display it as (7 nCr 4). To retrieve this operator, push the MATH button and check the PRB (probability) menu for nCr.

Ex2: What is the value of value 4 in row 7?

V_7,4 = (7 nCr 4) = 35

Method 3) Simpler Equation

Let p be the value of the entry immediately prior to our current entry in a row (p = V_n,k-1). This means that if we are evaluating V_6,3 then p represents the value V_6,2. Now we can use two different, simpler equations to determine values in a row.

We can find the value V_n,k with an easier equation provided the row is at least 4 (n>3) and index is at least 2 (k>1). The equation is V_n>3,k>1 = p[n-(k-1)]/k.

Ex3: Find V in the same triangle as from the first example above.

V_n,k = V_4,2 so p = V_4,1 = 4

V_4,2 = p[n-(k-1)]/k = (V_4,1)[4-(2-1)]/2 = 4(3)/2 = 6

We received 6, the same value as before and the same value used in the original triangle up top.

Method 4) Exponentiating the Magic 11

This method only works well for rows up to and including row 4. To find out the values for row 3 (n=3, "fourth" row), simply use your calculator to evaluate 11^3. This works on EVERY row and in EVERY base.

Ex4: Find the values in row 4 (n=4).

R_4 = 11^4 = 14641

Split these digits up into seperate values and we get "1 4 6 4 1" for row 4. Notice the 6 we've solved for with the last two methods is present as well!

Aside: The better application for the Magic 11 method is finding values for 11^n when you know what row n looks like in Pascal's Triangle. To go from row 8 to the value of 11^8 is not too bad.

R_8 = 1 8 28 56 70 56 28 8 1

We will ignore the first 1 and last three digits. This means we start off with 11^8 = 1...881

To fill it in, add adjacent pairs of numbers, starting after the first 1:

11^8 = 1 (8+2) ... 8 8 1

Because (8+2)=10, we need to increment the place to the left up by 1.

Now 11^8 = 2 0 (8+5) ... 8 8 1

Increment the 0 by 1 and leave the 3.

11^8 = 2 1 3 (6+7) ... 8 8 1

11^8 = 2 1 4 3 (0+5) ... 8 8 1 (Notice that (0+5) is less than 10, so we can quickly continue to the next pair)

11^8 = 2 1 4 3 5 (6+2) 8 8 1

11^8 = 214358881

EXAMPLE: Populate row 7 of Pascal's Triangle without the method by which you draw the entire structure, adding neighboring values to find the one below them. Hint: Remember to fill out the first two and last two values in a row by the method "1 n . . . n 1".

n=7

R_7 = 1 7 (V_2) (V_3) (V_4) (V_5) 7 1

V_2 = V_7,2 = n!/[1!(n-k)!] = 7!/[2!(7-2)!] = (7*6*5!)/(2!5!) = 42/2 = 21 (Method 1)

V_3 = V_7,3 = p[n-(k-1)]/k = 21(7-2)/3 = 35 (Method 3)

V_4 = 35(7-3)/4 = 35

V_5 = (7 nCr 5) = 21 (Method 2)

So R_7 = 1 7 21 35 35 21 7 1