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Pascal’s Triangle

Pascal’s Triangle. 1 1 1 2 1 3 3 1 1 4 6 4 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1. Pascal’s. Triangle. Each term is derived from the previous two terms above it. It is denoted  t n,r n  row number

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Pascal’s Triangle

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  1. Pascal’s Triangle

  2. 1 • 1 • 1 2 1 • 3 3 1 • 1 4 6 4 1 • 5 10 10 5 1 • 1 6 15 20 15 6 1 • 1 7 21 35 35 21 7 1 Pascal’s Triangle

  3. Each term is derived from the previous two terms above it. It is denoted  tn,r n  row number r  place number (counted from left side) Counting always starts with Zero (refer to page 247 of text)

  4. Create the equation to form each entry Tn,r = tn-1,r-1 + tn-1,r What does this mean? Remember that we start counting using the 0! Find the sum of the first 5 rows of the triangle → 2n

  5. The first 6 terms in row 25 are 1,25,300,2300,12650 and 53130. What are the first 6 terms in row 26? Use the formula T26,1 = t25,0 + t25,1 = 1 + 25 = 26 t26,2 = t25,1 + t25,2 = 25 + 300 = 325 t26,5 = t25,4 + t25,5 = 12650 + 53130 = 65780

  6. Show the formula that would give you T34,8 Which row in Pascal’s triangle has the sum of its terms equal to 32768? From our investigations before we know that the sum of each row is just 2n. So we just need to find out how many 2’s there are in 32768  15

  7. Divisibility Is ? Row 0  n/a Row 1  n/a Row 2  0.5 no Row 3  1 yes Row 4  1.5 no Row 5  2 yes Row 6  2.5 no Row 7  3 yes Therefore it is only divisible for odd-numbered rows.

  8. Equilateral Triangles 1, 3, 6, 10, 15, 21, 28, 36, 45, 55 These are the number of “balls” that are placed in an equilateral triangle 1 3 6 10

  9. It follows the pattern 1+2+3+4+5+6+7+8 Look at Pascal’s Triangle….where do the numbers fall that are in the pattern? They fall on the diagonal with the term tn+1,2 The term number is always 2, however the row is always 1 more than the number of columns in the equilateral. Example Find how many balls there are in 15 rows. T15+1,2 = t16,2

  10. Perfect Squares Is there a relationship between perfect squares and the sums of pairs of entries in Pascal’s triangle? • n n2 Entries in Pascal’s triangle Terms in the triangle • 1 1 t2,2 • 4 1 + 3 • 9 3 + 6 • 4 16 6 + 10 t2,2 + t3,2 t3,2 + t4,2 t4,2 + t5,2 Each perfect square greater than one is equal to the sum of a pair of adjacent terms on the third diagonal of Pascal’s triangle n2 = tn,2 + tn+1,2 for n>1

  11. Homework Pg 251 # 1, 2ac, 3bc, 4ac,5, 6a, 8

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