Daily math problems for August 6, 2026 — Sequences and series, Analytic Geometry, Algebra & more

$a_{20} = a_1 + 19d = 2 + 19 \cdot 3 = 2 + 57 = 59$.

$|AB| = 3$, $|AC| = 4$, $|BC| = \sqrt{9 + 16} = 5$. Perimeter $= 3 + 4 + 5 = 12$.

A $3$-cycle has sign $(-1)^{3-1} = +1$. (Equivalently, it can be written as the product of two transpositions: $(1\;2\;3) = (1\;3)(1\;2)$.)

$\overrightarrow{AB} = (4, 0)$ and $\overrightarrow{DC} = (4, 0)$ — same vector, so $ABCD$ is a parallelogram. Base $= 4$, height $= 3$. Area $= 12$.

$2\cos^2 x + \cos x - 1 = 0 \Rightarrow (2\cos x - 1)(\cos x + 1) = 0$. $\cos x = 1/2$: 2 solutions ($\pi/3, 5\pi/3$). $\cos x = -1$: 1 solution ($\pi$). Total: $3$.

Pair each subset $S$ with $S \triangle \{1\}$ (i.e., toggle the element $1$). This pairing has no fixed points and flips the parity of the sum, so even-sum and odd-sum subsets are equinumerous: $2^{10}/2 = 512$.

$R(\theta) R(-\theta) = I_2$, so $R(\theta)^{-1} = R(-\theta)$.

$n \ln\!\left(1 + \tfrac{1}{n^2}\right) \sim n \cdot \tfrac{1}{n^2} = \tfrac{1}{n} \to 0$, so the limit equals $e^0 = 1$.

Direct substitution gives $\tfrac{0}{0}$. Differentiate top and bottom: $\displaystyle\lim_{x\to 0}\dfrac{e^x - 1}{2x}$ — still $\tfrac{0}{0}$. Once more: $\displaystyle\lim_{x\to 0}\dfrac{e^x}{2} = \dfrac{1}{2}$.

$7^2 = 49$, $7^4 = 49^2 = 2401 \equiv 1 \pmod{100}$. Since $2024 = 4 \cdot 506$, we have $7^{2024} = (7^4)^{506} \equiv 1^{506} = 1 \pmod{100}$. Last two digits: $01$.