# Tag Info

0

Consider a combinatorial argument. Let $\Lambda = \{a, b, c\}$ be our alphabet. We select the $i$ positions in an $n$ letter word for $\{a, b\}$ characters in $\binom{n}{i}$ ways and multiply by $2^{i}$ as we have a word (is each selected slot a $a$ or is it a $b$?). The remaining $n-i$ slots are all $c$'s. Now we add up over all possible values of $i$: ...

3

we have $$\left(1 + x\right)^n = 1 + {n \choose 1} x + {n \choose 2} x^2+ \cdots + x^n \tag 1$$ now putting $x = 2$ in $(1)$ gives you the result.

4

In the binomial theorem, $$(x+y)^n={n \choose 0}x^{n-0}y^0+{n \choose 1}x^{n-1}y^1+{n \choose 2}x^{n-2}y^2+\cdots+{n \choose n}x^{n-n}y^n=\sum_{i=1}^{n} {n \choose i} x^{n-i}y^i$$ Put $x=1$ and $y=2$.

0

Sometimes we have asymptotics with most significant term $M \to \infty$ we do this $$\log(M+A) = \log(M\cdot(1+S))$$ so that $S=A/M$ is "small" in the sense $S=o(1)$, and then $$\log(M\cdot(1+S)) = \log M + \log(1+S) = \log M + S - \frac{1}{2}S^2+\frac{1}{3}S^3+\cdots$$ reference G. A. Edgar, Transseries for beginners, Real Anal. Exchange 35 (2010), no. ...

1

A binary tree of height $n$ can hold $1+2+4+...2^{n-1}=2^n - 1$ elements. This is because if $s= 1+2+4+...2^{n-1}$ then $2s=2+4+8+...2^n$ so when you subtract $s$ from $2s$ everything except the $2^n$ and the $1$ cancel to give $$2s-s=s=2^n-1$$

3

As it seems to be a multiple choice question, we can find the answer without much computing: all factors have the form $$(2n)^4+324=4(4n^4+81).$$ Since there are as many factors in the numerator as in the denominator, the $4$s cancel out, which results in a fraction with odd numerator and denominator, hence this fraction must simplify to an odd number. ...

2

Using the following identity: $$a^4 + 4\cdot 3^4 = (a^2 + 2 \cdot 3^2 - 2\cdot 3\cdot a)(a^2 + 2 \cdot 3^2 + 2\cdot 3\cdot a) = (a(a-6) + 18)(a(a+6)+18)$$ Most of the terms cancel out and you are left with: $$\frac{58(64)+18}{4(-2)+18} = \frac{3730}{10} = 373$$ As KprimeX mentioned, this flows from the Sophie Germain Identity.

4

Hint You may try with the Sophie Germain identity. As $(10^4+324)=(10^4+4\times3^4)$

1

Yes. Bounds for such sums are known more generally for Laurent polynomials. It is a useful (but lengthy) exercise to derive the bound $$\left|\sum_{x\in\Bbb{F}_q^*}\psi( f(x)+g(\frac1x))\right|\le (\deg f+\deg g)\sqrt q$$ with the method described in Lidl & Niederreiter. Here $f$ and $g$ can be any polynomials that cannot be written in the form ...

1

For simplicity, consider the case $x_i$ are sorted in descending order. $$x_1 \ge x_2 \ge x_3 \ge \cdots$$ Let $M = x_1 = \max\limits_{1\le i \le n}\{ x_i \}$, we have $$\log \sum\limits_{i=1}^n e^{x_i} = M + \log\sum\limits_{i=1}^n e^{x_i-M} = M + \log\left(1 + \sum_{i=2}^n e^{x_i-M}\right)$$ For any $\Lambda > 0$, let $m$ be a index such that M = ...

Top 50 recent answers are included