 9.4.1: 2 + 4 + 6 + + 2n = n1n + 12
 9.4.2: 3 + 5 + 7 + + 12n + 12 = n1n + 22
 9.4.3: 3 + 4 + 5 + + 1n + 22 = 1 2 n1n + 52
 9.4.4: 3 + 5 + 7 + + 12n + 12 = n1n + 22
 9.4.5: 2 + 5 + 8 + + 13n  12 = 1 2 n13n + 12
 9.4.6: 1 + 4 + 7 + + 13n  22 = 1 2 n13n  12
 9.4.7: 1 + 2 + 22 + + 2n1 = 2n  1
 9.4.8: 1 + 3 + 32 + + 3n1 = 1 2 13n  12
 9.4.9: 1 + 4 + 42 + + 4n1 = 1 3 14n  12
 9.4.10: 1 + 5 + 52 + + 5n1 = 1 4 15n  12
 9.4.11: 1 1 # 2 + 1 2 # 3 + 1 3 # 4 + + 1 n1n + 12 = n n + 1
 9.4.12: 1 1 # 3 + 1 3 # 5 + 1 5 # 7 + + 1 12n  1212n + 12 = n 2n + 1
 9.4.13: 12 + 22 + 32 + + n2 = 1 6 n1n + 1212n + 12
 9.4.14: 13 + 23 + 33 + + n3 = 1 4 n2 1n + 122
 9.4.15: 4 + 3 + 2 + + 15  n2 = 1 2 n19  n2
 9.4.16: 2  3  4   1n + 12 =  1 2 n1n + 32
 9.4.17: 1 # 2 + 2 # 3 + 3 # 4 + + n1n + 12 = 1 3 n1n + 121n + 22
 9.4.18: 1 # 2 + 3 # 4 + 5 # 6 + + 12n  1212n2 = 1 3 n1n + 1214n  12
 9.4.19: n is divisible by 2
 9.4.20: n is divisible by 3.
 9.4.21: n is divisible by 2.
 9.4.22: n1n + 121n + 22 is divisible by 6.
 9.4.23: If then xn x 7 1, 7 1.
 9.4.24: If then xn 0 6 x 6 1, 6 1.
 9.4.25: a  b is a factor of an  bn
 9.4.26: a + b is a factor of a2n+1 + b2n+1
 9.4.27: (1 + a) n 1 + na, for a 7 0
 9.4.28: Show that the statement is a prime number is true for but is not tr...
 9.4.29: Show that the formula 2 + 4 + 6 + + 2n = n2 + n + 2 obeys Condition...
 9.4.30: Use mathematical induction to prove that if then a + ar + ar2 + + a...
 9.4.31: Use mathematical induction to prove that a + 1a + d2 + 1a + 2d2 + +...
 9.4.32: Extended Principle of Mathematical Induction The Extended Principle...
 9.4.33: Geometry Use the Extended Principle of Mathematical Induction to sh...
 9.4.34: How would you explain the Principle of Mathematical Induction to a ...
Solutions for Chapter 9.4: Mathematical Induction
Full solutions for College Algebra  9th Edition
ISBN: 9780321716811
Solutions for Chapter 9.4: Mathematical Induction
Get Full SolutionsSince 34 problems in chapter 9.4: Mathematical Induction have been answered, more than 7872 students have viewed full stepbystep solutions from this chapter. This textbook survival guide was created for the textbook: College Algebra, edition: 9. This expansive textbook survival guide covers the following chapters and their solutions. Chapter 9.4: Mathematical Induction includes 34 full stepbystep solutions. College Algebra was written by and is associated to the ISBN: 9780321716811.

Diagonalizable matrix A.
Must have n independent eigenvectors (in the columns of S; automatic with n different eigenvalues). Then SI AS = A = eigenvalue matrix.

Echelon matrix U.
The first nonzero entry (the pivot) in each row comes in a later column than the pivot in the previous row. All zero rows come last.

Elimination matrix = Elementary matrix Eij.
The identity matrix with an extra eij in the i, j entry (i # j). Then Eij A subtracts eij times row j of A from row i.

Exponential eAt = I + At + (At)2 12! + ...
has derivative AeAt; eAt u(O) solves u' = Au.

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.

Free variable Xi.
Column i has no pivot in elimination. We can give the n  r free variables any values, then Ax = b determines the r pivot variables (if solvable!).

Inverse matrix AI.
Square matrix with AI A = I and AAl = I. No inverse if det A = 0 and rank(A) < n and Ax = 0 for a nonzero vector x. The inverses of AB and AT are B1 AI and (AI)T. Cofactor formula (Al)ij = Cji! detA.

Kirchhoff's Laws.
Current Law: net current (in minus out) is zero at each node. Voltage Law: Potential differences (voltage drops) add to zero around any closed loop.

Krylov subspace Kj(A, b).
The subspace spanned by b, Ab, ... , AjIb. Numerical methods approximate A I b by x j with residual b  Ax j in this subspace. A good basis for K j requires only multiplication by A at each step.

Left inverse A+.
If A has full column rank n, then A+ = (AT A)I AT has A+ A = In.

Network.
A directed graph that has constants Cl, ... , Cm associated with the edges.

Projection p = a(aTblaTa) onto the line through a.
P = aaT laTa has rank l.

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.

Right inverse A+.
If A has full row rank m, then A+ = AT(AAT)l has AA+ = 1m.

Semidefinite matrix A.
(Positive) semidefinite: all x T Ax > 0, all A > 0; A = any RT R.

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

Symmetric factorizations A = LDLT and A = QAQT.
Signs in A = signs in D.

Triangle inequality II u + v II < II u II + II v II.
For matrix norms II A + B II < II A II + II B IIĀ·

Unitary matrix UH = U T = UI.
Orthonormal columns (complex analog of Q).

Vector space V.
Set of vectors such that all combinations cv + d w remain within V. Eight required rules are given in Section 3.1 for scalars c, d and vectors v, w.
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