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The binomial theorem givesThe central binomial coefficient is the largest of these $2n+1$ nonnegative binomial coefficients, so

For the upper bound, fix a prime number $p\le 2n$. By Legendre formula,Each summand is either zero or one. Hence the P-adic valuation satisfiesIf $p\le \sqrt{2n}$, this impliesThere are at most $\sqrt{2n}$ such prime numbers. If instead $p>\sqrt{2n}$, then $p^2>2n$, so the P-adic valuation is at most one. Multiplying the prime-power contributions therefore yields

Because $k$ is a perfect field, for each $a\in k$ choose a compatible sequenceand choose arbitrary lifts $x_n\in {\mathcal{O}}_K$ of $a_n$. If $u\equiv v(\mathrm{mod}{\pi }^m)$, the binomial theorem and the fact that the residue characteristic is $p$ giveIt follows that $x_{n+1}^{p^{n+1}}\equiv x_n^{p^n}(\mathrm{mod}{\pi }^{n+1})$. Thus $(x_n^{p^n})$ is Cauchy, and completeness definesThe same congruence shows that the limit is independent of all lift choices. Taking products before passing to the limit proves $[ab]=[a][b]$, and reduction gives $[a]\equiv a(\mathrm{mod}\pi )$.

For uniqueness, let $s,t:k\to {\mathcal{O}}_K$ be two multiplicative lifts. Given $a$ and any $n$, choose $b\in k$ with $b^{p^n}=a$. Since $s(b)\equiv t(b)(\mathrm{mod}\pi )$, repeated powering yieldsCompleteness and separation force $s(a)=t(a)$. This is the unique Teichmuller lift.

For $x\in {\mathcal{O}}_K$, let $a_0$ be its residue and put $x_1=(x-[a_0])/\pi$. Repeat with $x_1,x_2,\dots$. Induction givesThe remainder tends to zero, proving the Teichmuller expansionReduction after subtracting successive partial sums also proves uniqueness of the digits.

The binomial theorem gives $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)\le {\sum }_{k=0}^{2n}\left(\genfrac{}{}{0ex}{}{2n}{k}\right)=4^n$. ThereforeThe geometric series with ratio $4/9$ converges. By the comparison test, the given series is absolutely convergent.

Expand $(I+B{)}^2$ in components:If $B^2=0$, all powers $B^k$ for $k\ge 2$ vanish. The matrix binomial theorem, valid here because $I$ commutes with $B$, therefore yieldsThese powers of the identity plus a square-zero matrix can also be proved by induction: $(I+nB)(I+B)=I+(n+1)B$. For the displayed example, $B^2=0$ directly, andThis is the repeated function composition of a shear mapping. The argument gives every nonnegative integer $n$; since $(I+B{)}^{-1}=I-B$, the same formula also holds for negative integers.

The number $R$ of successes has a binomial distribution. Each arrangement of $r$ successes and $n-r$ failures has probability $p^r{q}^{n-r}$ by independence, and the binomial coefficient counts the arrangements. ThereforeThe final equality is the binomial theorem. Writing $R$ as a sum of independent Bernoulli random variables gives its expectation and variance: $\mathbb{E}R=np$, $\mathrm{Var}(R)=npq$. Thus the matching normal distribution hasThe normal approximation is appropriate when both expected counts $np$ and $nq$ are sufficiently large; large $n$ alone is insufficient if $p$ is extremely close to zero or one.

For the guessing student, $n=60$, $p=1/4$, $q=3/4$. Therefore the expected number of correct answers is $\boxed{15}$, and ${\sigma }^2=45/4$, $\sigma =3\sqrt{5}/2$. The inclusive integer range $13,14,15,16,17$ becomes $12.5\le X\le 17.5$ after the continuity correction. Standardizing the normal distribution givesHere $(5/2)/(3\sqrt{5}/2)=\sqrt{5}/3$, and the last step uses the supplied bound. The approximate probability is $0.54394$. The normal approximation provides the requested conclusion; as a separate check, summing the five exact binomial probabilities gives approximately $0.54370$, also exceeding one half.

For the binomial distribution and $T\ge 1$, the identity $s\left(\genfrac{}{}{0ex}{}{T}{s}\right)=T\left(\genfrac{}{}{0ex}{}{T-1}{s-1}\right)$ gives the expectationThe final sum is one by the binomial theorem. For $T=0$ the variable is identically zero, so the same answer holds:

Set $z=e^{i\theta }$, so $2i\sin \theta =z-z^{-1}$. The binomial theorem givesUse $z^k-z^{-k}=2i\sin (k\theta )$ and $(2i{)}^5=32i$ to obtainThis is a finite Fourier sine series, with only odd harmonics.

The binomial theorem assigns binomial coefficient $\left(\genfrac{}{}{0ex}{}{9}{3}\right)$ to $x^3$. Since $\left(\genfrac{}{}{0ex}{}{9}{3}\right)=9\cdot 8\cdot 7/(3\cdot 2\cdot 1)=84$, the requested term is $84x^3$.

The sum is finite, so differentiation passes through it without a convergence issue. Each lower limit $n$ is constant with respect to $x$, and the fundamental theorem of calculus gives the same upper-end integrand. Hence the derivative isThe binomial theorem at $1+1$ gives the sum $2^N$. Therefore

The binomial theorem gives coefficients $1,4,6,4,1$ for the fourth power. Substituting $2x$ for its variable gives

By De Moivre's theorem, $\cos 5\theta$ is the real part of $(\cos \theta +i\sin \theta {)}^5$. Writing $c=\cos \theta$ and $s=\sin \theta$, the binomial theorem givesUse the Pythagorean trigonometric identity $s^2=1-c^2$ and expand:Therefore

The binomial theorem givesThus, to three decimal places,The first three nonconstant terms already determine this rounding; the remaining positive terms total less than $1.3\times 10^{-6}$.

By De Moivre's theorem, $\sin 5\theta$ is the imaginary part of $(\cos \theta +i\sin \theta {)}^5$. The binomial theorem gives, with $s=\sin \theta$,Use ${\cos }^2\theta =1-s^2$ and simplify:The requested equation cancels the $16s^5$ term, leaving $5s(1-4s^2)=0$. Thus $s=0$ or $s=\pm 1/2$. The complete list in $0\le \theta <2\pi$ is

Factor the leading term out of the Maclaurin series for the sine:The binomial theorem gives, through degree four inside the bracket,The coefficient of $x^4$ is $1/20+5/12=7/15$. Multiplying back by $x^6$ yields

The binomial theorem gives $(1+x{)}^8={\sum }_{k=0}^8(\genfrac{}{}{0ex}{}{8}{k})x^k$. The required binomial coefficients are thereforeHere $[x^k]$ denotes the coefficient of $x^k$ in the polynomial.

For $0\le t\le 1$, the triangle inequality givesTaking the supremum proves $\boxed{d(p_n,p)\le \rho (p_n,p)}$. Hence convergence for $\rho$ implies convergence for $d$.

The converse is false. For $n\ge 1$, take $q_n(t)=(1-t{)}^n/n$ and $p=0$. Its supremum norm is $d(q_n,0)=1/n\to 0$, whereas the binomial theorem givesThe alternating coefficients cancel on $[0,1]$; the polynomial coefficient sum norm measures their absolute values before any cancellation.

In field characteristic $p$, the binomial theorem gives $(X-b{)}^p=X^p-b^p$, since the intermediate binomial coefficients are divisible by the prime number $p$. If $b\in F$ is a root of $X^p-a$, then $b^p=a$, and hence

Independence and convolution giveby the binomial theorem. Thus

In field characteristic $p$, the binomial theorem gives $(X-b{)}^p=X^p-b^p$, since the intermediate binomial coefficients are divisible by the prime number $p$. If $b\in F$ is a root of $X^p-a$, then $b^p=a$, and hence

Independence and convolution giveby the binomial theorem. Thus