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

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

$\mathrm{\mid }AB\mathrm{\mid }=3$, $\mathrm{\mid }AC\mathrm{\mid }=4$, $\mathrm{\mid }BC\mathrm{\mid }=\sqrt{9+16}=5$. 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: $(123)=(13)(12)$.)

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

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

Pair each subset $S$ with $S\mathrm{△}\{1\}$ (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: $2^{10}\mathrm{/}2=512$.

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

$n\ln \text{\hspace{0.17em}⁣}(1+\frac{1}{n^2})\sim n\cdot \frac{1}{n^2}=\frac{1}{n}\to 0$, so the limit equals $e^0=1$.

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

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