 11.4.1: The first step in proving a formula by ________ ________ is to show...
 11.4.2: The ________ differences of a sequence are found by subtracting con...
 11.4.3: A sequence is an ________ sequence if the first differences are all...
 11.4.4: If the ________ differences of a sequence are all the same nonzero ...
 11.4.5:
 11.4.6: Pk
 11.4.7: P 2k 1 k k 2 k 3 2 6
 11.4.8: Pk k 3 P 2k 1
 11.4.9: Pk 3 k 2 k 3
 11.4.10: Pk k2 2 k 12
 11.4.11: 2 4 6 8 . . . 2n n n 1
 11.4.12: 3 7 11 15 . . . 4n 1 n 2n 1 2
 11.4.13: 2 7 12 17 . . . 5n 3 n 2 5n 1 3
 11.4.14: 1 4 7 10 . . . 3n 2 n 2 3n 1 2
 11.4.15: 1 2 22 23 . . . 2n1 2n 1 1
 11.4.16: 2 1 3 32 33 . . . 3n1 3n 1 1
 11.4.17: 1 2 3 4 . . . n n n 1 2
 11.4.18: 13 23 33 43 . . . n3 n2 n 1 2 4
 11.4.19: 12 32 52 . . . 2n 12 n 2n 1 2n 1 3 1
 11.4.20: 1 1 11 1 21 1 3 . . . 1 1 n n 1
 11.4.21: n i1 i5 n2 n 12 2n2 2n 1 12
 11.4.22: n i1 i 4 n n 1 2n 1 3n2 3n 1 30
 11.4.23: n i1 i i 1 n n 1 n 2 3
 11.4.24: n i1 1 2i 1 2i 1 n 2n 1
 11.4.25: n! >
 11.4.26: 4 3
 11.4.27: 1 1 2 1 3 . . . 1 n > n, n 2 4
 11.4.28: x y n 1 < x y n ,
 11.4.29: 1 a n 1 a > 0
 11.4.30: 2n n 3
 11.4.31: abn an bn
 11.4.32: a b n an bn
 11.4.33: If then
 11.4.34: If then
 11.4.35: Generalized Distributive Law:
 11.4.36: a bi
 11.4.37: A factor of is 3.
 11.4.38: A factor of is 3.
 11.4.39: A factor of is 2.
 11.4.40: A factor of is 3.
 11.4.41: A factor of is 5.
 11.4.42: A factor of is 5.
 11.4.43: 1, 5, 9, 13, . . .
 11.4.44: 25, 22, 19, 16, . . .
 11.4.45: 1, , . . . 9 10, 81 100, 729 1000, . .
 11.4.46: 3, 9 2, 27 4 , 81 8 1, , . . . 9
 11.4.47: 1 4 , 1 12, 1 24, 1 40, . . . , 1 2n n 1 , . . .
 11.4.48: 2 3 , 1 3 4 , 1 4 5 , 1 5 6 , . . . ,
 11.4.49: 15 n1 n
 11.4.50: 30 n1 n
 11.4.51: 6 n1 n2
 11.4.52: 10 n1 n3
 11.4.53: 5 n1 n4
 11.4.54: 8 n1 n5
 11.4.55: 6 n1 n2 n
 11.4.56: 20 n1 n 3 n
 11.4.57: 6 i1 6i 8i 3
 11.4.58: 10 j1 3 1 2 j 1 2 j 2 6
 11.4.59: 5, 13, 21, 29, 37, 45, . . .
 11.4.60: 2, 9, 16, 23, 30, 37, . . .
 11.4.61: 6, 15, 30, 51, 78, 111, . . .
 11.4.62: 0, 6, 16, 30, 48, 70, . . .
 11.4.63: 2, 1, 6, 13, 22, 33, . . .
 11.4.64: 1, 8, 23, 44, 71, 104, . . .
 11.4.65: a1 0
 11.4.66: a1 2
 11.4.67: a1 3
 11.4.68: a2 3
 11.4.69: a0 2
 11.4.70: a0 0
 11.4.71: a1 2
 11.4.72: a1 0
 11.4.73: a0 3, a1 3, a4 15
 11.4.74: a0 7, a1 6, a3 10
 11.4.75: a0 3, a2 1, a4 9 a0 7, a1 6, a3 10
 11.4.76: a0 3, a2 0, a6 36
 11.4.77: a1 0, a2 8, a4 30
 11.4.78: a0 3, a2 5, a6 57 a
 11.4.79: DATA ANALYSIS: RESIDENTS The table shows the numbers (in thousands)...
 11.4.80: CAPSTONE In your own words,
 11.4.81: If the statement is true but the true statement does not imply that...
 11.4.82: If the statement is true and implies then is also true.
 11.4.83: If the second differences of a sequence are all zero, then the sequ...
 11.4.84: A sequence with terms has second differences
 11.4.85: If a sequence is arithmetic, then the first differences of the sequ...
Solutions for Chapter 11.4: Mathematical Induction
Full solutions for Algebra and Trigonometry  8th Edition
ISBN: 9781439048474
Solutions for Chapter 11.4: Mathematical Induction
Get Full SolutionsThis expansive textbook survival guide covers the following chapters and their solutions. Since 85 problems in chapter 11.4: Mathematical Induction have been answered, more than 48862 students have viewed full stepbystep solutions from this chapter. This textbook survival guide was created for the textbook: Algebra and Trigonometry, edition: 8. Algebra and Trigonometry was written by and is associated to the ISBN: 9781439048474. Chapter 11.4: Mathematical Induction includes 85 full stepbystep solutions.

Associative Law (AB)C = A(BC).
Parentheses can be removed to leave ABC.

Cholesky factorization
A = CTC = (L.J]))(L.J]))T for positive definite A.

Determinant IAI = det(A).
Defined by det I = 1, sign reversal for row exchange, and linearity in each row. Then IAI = 0 when A is singular. Also IABI = IAIIBI and

Distributive Law
A(B + C) = AB + AC. Add then multiply, or mUltiply then add.

Factorization
A = L U. If elimination takes A to U without row exchanges, then the lower triangular L with multipliers eij (and eii = 1) brings U back to A.

Four Fundamental Subspaces C (A), N (A), C (AT), N (AT).
Use AT for complex A.

Free columns of A.
Columns without pivots; these are combinations of earlier columns.

Full column rank r = n.
Independent columns, N(A) = {O}, no free variables.

Hermitian matrix A H = AT = A.
Complex analog a j i = aU of a symmetric matrix.

Indefinite matrix.
A symmetric matrix with eigenvalues of both signs (+ and  ).

Least squares solution X.
The vector x that minimizes the error lie 112 solves AT Ax = ATb. Then e = b  Ax is orthogonal to all columns of A.

Left nullspace N (AT).
Nullspace of AT = "left nullspace" of A because y T A = OT.

Markov matrix M.
All mij > 0 and each column sum is 1. Largest eigenvalue A = 1. If mij > 0, the columns of Mk approach the steady state eigenvector M s = s > O.

Nullspace matrix N.
The columns of N are the n  r special solutions to As = O.

Nullspace N (A)
= All solutions to Ax = O. Dimension n  r = (# columns)  rank.

Rank r (A)
= number of pivots = dimension of column space = dimension of row space.

Reflection matrix (Householder) Q = I 2uuT.
Unit vector u is reflected to Qu = u. All x intheplanemirroruTx = o have Qx = x. Notice QT = Q1 = Q.

Simplex method for linear programming.
The minimum cost vector x * is found by moving from comer to lower cost comer along the edges of the feasible set (where the constraints Ax = b and x > 0 are satisfied). Minimum cost at a comer!

Spanning set.
Combinations of VI, ... ,Vm fill the space. The columns of A span C (A)!

Toeplitz matrix.
Constant down each diagonal = timeinvariant (shiftinvariant) filter.