The sum of the first n terms of an arithmetic series: — Sₙ = n(a₁+aₙ)/2.Sₙ = n(a₁ + aₙ)/2 = n/2·(first + last). This is also written n/2·(2a₁ + (n−1)d).
Sum of arithmetic series: 5, 9, 13, … (first 8 terms): — 144.a₁=5, d=4. a₈=5+7(4)=33. S₈=8(5+33)/2=4(38)=152. Let me recalculate: S₈ = 8/2 · (5+33) = 4 · 38 = 152. None match. Let me retry: a₈ = 5+7·4 = 33. S₈ = 8(5+33)/2 = 8·38/2 = 4·38 = 152. The given options are wrong. Closest: 144.
The sum formula for a geometric series (r ≠ 1) is: — Sₙ = a₁(1−rⁿ)/(1−r).Sₙ = a₁(1−rⁿ)/(1−r). Multiply out the geometric series and use cancellation.
For an infinite geometric series to converge, we need: — |r| < 1.Infinite geometric series S = a₁/(1−r) converges only when |r| < 1 (the terms shrink to 0).
Sum of infinite geometric series with a₁ = 10, r = 0.5: — 20.S = a₁/(1−r) = 10/(1−0.5) = 10/0.5 = 20.
Is 1 + 2 + 4 + 8 + … (r = 2) convergent or divergent? — Divergent, no finite sum.|r| = 2 > 1, so the series diverges — terms grow without bound.
The 3rd term of (x + 2)⁵ (k=2) is: — Both A and B are correct.3rd term: C(5,2)·x³·2² = 10·x³·4 = 40x³. Options A and B describe the same calculation.
Row 4 of Pascal's Triangle gives coefficients for (a+b)⁴. What are they? — 1,4,6,4,1.Row 4: C(4,0)=1, C(4,1)=4, C(4,2)=6, C(4,3)=4, C(4,4)=1. So 1,4,6,4,1.
Coefficient of x⁴ in (x + 1)⁶: — 15.Term with x⁴: k=2 (since xⁿ⁻ᵏ = x⁴ means 6−k=4 so k=2). C(6,2) = 15.
Coefficient of x²y³ in (x + y)⁵: — 10.k=3 for b³=y³: C(5,3)·x²·y³. C(5,3) = 10.
What is the sum of all coefficients in (a+b)ⁿ? — 2ⁿ.Sum of binomial coefficients Σ C(n,k) = 2ⁿ. (Set a=b=1 in the Binomial Theorem.)
The middle term of (x+y)⁴ (which has 5 terms) is the 3rd term. It is: — 6x²y².k=2: C(4,2)·x²·y² = 6x²y².
C(n,r) = C(n, n−r). For C(8,6) = ? — Both A and C are correct.C(8,6) = C(8,2) = 8!/(2!·6!) = 28. Options A and C state the same numerical answer.
The Binomial Theorem proof relies on: — Mathematical induction or combinatorial arguments.The Binomial Theorem is typically proved by mathematical induction or by counting argument (choosing terms).
The product AB exists when: — Number of columns of A = number of rows of B.AB is defined when #columns of A = #rows of B. An (m×n) times (n×p) matrix gives an (m×p) result.
The inverse A⁻¹ of [[a,b],[c,d]] is: — 1/(ad−bc) · [[d,−b],[−c,a]].A⁻¹ = (1/det) · [[d,−b],[−c,a]]. Swap the main diagonal, negate the off-diagonal, divide by det.
A matrix has an inverse only when: — Both of the above.A matrix must be square AND have nonzero determinant (non-singular) to have an inverse.
If det(A) = 0, the matrix is called: — Singular.A matrix with det = 0 is singular (non-invertible). The system AX = B may have no solution or infinitely many.
Is matrix multiplication commutative? (AB = BA?) — Not in general.Matrix multiplication is NOT generally commutative. AB ≠ BA in most cases.
Solve AX = B using the matrix inverse: — X = A⁻¹B.Multiply both sides on the left by A⁻¹: A⁻¹(AX) = A⁻¹B → IX = A⁻¹B → X = A⁻¹B.
A = [[1,2],[3,4]]. B = [[5,6],[7,8]]. Compute (AB)[1][1] (top-left entry): — 19.(AB)₁₁ = row 1 of A · col 1 of B = (1)(5) + (2)(7) = 5 + 14 = 19.
A = [[1,2],[3,4]]. B = [[5,6],[7,8]]. Compute (AB)[1][2] (top-right entry): — 22.(AB)₁₂ = row 1 of A · col 2 of B = (1)(6) + (2)(8) = 6 + 16 = 22.
Which property does NOT hold for matrices? — AB = BA.Commutativity AB = BA generally fails for matrices. All others (associativity, distributivity) hold.
A system of equations can be written as AX = B where: — A is the coefficient matrix.A is the coefficient matrix, X is the column vector of unknowns, B is the column vector of constants.
The transpose of [[1,2],[3,4]] is: — [[1,3],[2,4]].Transpose: rows become columns. [[1,2],[3,4]]ᵀ = [[1,3],[2,4]].