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Binomial Identities

Binomial Identities. Expansion of (a + x) n. (a + x) = a + x = 1 C 0 a + 1 C 1 x (a + x)(a + x) = aa + ax + xa + xx = x 2 + 2ax + a 2 = 2 C 0 x 2 + 2 C 1 ax + 2 C 2 a 2 The 4 red terms are the “formal” expansion of (a+x) 2

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Binomial Identities

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  1. Binomial Identities

  2. Expansion of (a + x)n • (a + x) = a + x = 1C0a + 1C1x • (a + x)(a + x) = aa + ax + xa + xx = x2 + 2ax + a2 = 2C0x2 + 2C1ax + 2C2a2 • The 4 red terms are the “formal” expansion of (a+x)2 • The 3 blue terms are the “simplified” expansion of (a+x)2 • (a + x)(a + x)(a + x) = aaa + aax + axa + axx + xaa + xax + xxa + xxx = x3 + 3a2x + 3ax2 + a3 = 3C0x3 + 3C1a2x + 3C2ax2 + 3C3a3

  3. Generalizing . . . • In (a + x)4, how many terms does the: • formal expansion have? • simplified expansion have? • In (a + x)n, how many terms does the: • formal expansion have? • simplified expansion have?

  4. The Coefficient on akxn-k • The coefficient on akxn-kis the number of terms in the formal expansion that have exactly kas (and hence exactly n-kxs). • It is equal to the number of ways of choosing an a from exactly k of the n binomial factors: nCk.

  5. Binomial Theorem • (1 + x)n = nC0x0 + nC1x1 + nC2x2 + . . . nCnxn • In addition to the combinatorial argument that the coefficient of xi is nCi, we can prove this theorem by induction on n.

  6. Binomial Identities • nCk = n!/[k!(n-k)!] = nCn-k The number of ways to pick k objects from n = the ways to pick not pick k (i.e., to pick n-k). • Pascal’s identity: nCk = n-1Ck + n-1Ck-1 • The number of ways to pick k objects from n can be partitioned into 2 parts: • Those that exclude a particular object i: n-1Ck • Those that include object i: n-1Ck-1 • Give an algebraic proof of this identity.

  7. nCkkCm = nCmn-mCk-m • Each side of the equation counts the number of k-subsets with an m-subsubset. • The LHS counts: 1. Pick k objects from n: nCk 2. Pick m special objects from the k: kCm • The RHS counts: 1. Pick m special objects that will be part of the k-subset: nCm 2. Pick the k-m non-special objects: n-mCk-m

  8. Pascal’s Triangle • kth number in row n is nCk: k = 0 1 n = 0 k = 1 k = 2 n = 1 1 1 k = 3 n = 2 1 2 1 k = 4 1 3 3 1 n = 3 n = 4 1 4 6 4 1

  9. Displaying Pascal’s Identity k = 0 0C0 n = 0 k = 1 1C0 1C1 k = 2 n = 1 2C0 2C1 2C2 k = 3 n = 2 k = 4 3C0 3C1 3C2 3C3 n = 3 4C0 4C1 4C2 4C3 4C4 n = 4

  10. Block-walking Interpretation nCk = # strings of n Ls & Rs with k Rs. nCk = # ways to get to corner n,k starting from 0,0 k = 0 0C0 n = 0 k = 1 1C0 1C1 k = 2 n = 1 2C0 2C1 2C2 k = 3 n = 2 k = 4 3C0 3C1 3C2 3C3 n = 3 4C0 4C1 4C2 4C3 4C4 n = 4

  11. Pascal’s Identity via Block-walking # routes to corner n,k = # routes thru corner n-1,k + # routes thru corner n-1,k-1 k = 0 0C0 n = 0 k = 1 1C0 1C1 k = 2 n = 1 2C0 2C1 2C2 k = 3 n = 2 k = 4 3C0 3C1 3C2 3C3 n = 3 4C0 4C1 4C2 4C3 4C4 n = 4

  12. nC0 + nC1 + nC2 + . . . + nCn = 2n • LHS counts # subsets of n elements using the sum rule: partitioning the subsets according to their size (k value). • RHS counts # subsets of n elements using the product rule: • Is element 1 in subset? (2 choices) • Is element 2 in subset? (2 choices) … • Is element n in subset? (2 choices)

  13. rCr + r+1Cr + r+2Cr + . . . + nCr = n+1Cr+1 k = 0 0C0 n = 0 k = 1 1C0 1C1 k = 2 n = 1 2C0 2C1 2C2 k = 3 n = 2 k = 4 3C0 3C1 3C2 3C3 n = 3 4C0 4C1 4C2 4C3 4C4 n = 4

  14. rCr + r+1Cr + r+2Cr + . . . + nCr = n+1Cr+1 • RHS = routes to corner 4,2 • LHS: Partition the routes to 4,2 into those: • whose last right branch is at corner 1,1: 1C1 • whose last right branch is at corner 2,1: 2C1 • whose last right branch is at corner 3,1: 3C1 • For each subset of routes, there is only 1 way to complete the route from that corner to 4,2: RLL, RL, & R respectively. • The identity generalizes this argument.

  15. nC02 + nC12 + nC22 + … + nCn2 = 2nCn k = 0 0C0 n = 0 k = 1 1C0 1C1 k = 2 n = 1 2C0 2C1 2C2 k = 3 n = 2 k = 4 3C0 3C1 3C2 3C3 n = 3 4C0 4C1 4C2 4C3 4C4 n = 4

  16. nC02 + nC12 + nC22 + … + nCn2 = 2nCn • RHS = all routes to corner 4,2 • LHS partitions routes to 4,2 into those that: • go thru corner 2,0: 2C0 2C2 • go thru corner 2,1: 2C1 2C1 • go thru corner 2,2: 2C2 2C0 • The identity generalizes this argument: • # routes to 2n, n that go thru n,k = nCk nCn-k • Sum over k = 0 to n

  17. 123 + 234 + 345 +…+ (n-2)(n-2)n = ? • The general term = (k-2)(k-1)k = P(k,3) = k!/(k-3)! = 3! kC3 • Sum = 3!3C3 + 3!4C3 + 3!5C3 +...+ 3!nC3 = 3! [3C3 + 4C3 + 5C3 +...+ nC3 ] = 3!  n+1C4

  18. A Strategy • When the general term of a sum is not a binomial coefficient: • break it into a sum of P(n, k) terms, if possible; • rewrite these terms using binomial coefficients

  19. 12 + 22 + 32 +. . . + n2 = ? • General term: = k2 = k(k-1) + k = P(k, 2) + k = 2!kC2 + k

  20. Sum Σk=1,n(2!kC2 + k) = 2!Σk=1,nkC2+Σk=1,nk = 2!n+1C3 +n+1C2

  21. Another Strategy: Manipulate the Binomial Theorem • (1 + 1)n = 2n = nC0 + nC1 + . . . + nCn • (1 - 1)n = 0 = nC0 - nC1 + nC2 - . . . +(-1)nnCn or nC0 + nC2 . . . = nC1 + nC3 + . . . = 2n-1 • Differentiate the Binomial theorem, • n(1 + x)n-1 = 1nC1x0 + 2nC2x1 + 3nC3x2 + … + nnCnxn-1 • n(1 + 1)n-1 = 1nC1+ 2nC2+ 3nC3+ … + nnCn

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