What is the formula for doing this type of problems?
33439
Note: Problem from "Kenneth Rosen's DM and it's applications" and solution from "Students solution guide for use with ... applications"  Let P(n) be the number of strings not containg two containing two consecutive zeros or two consecutive ones.  I got the answer  P(n)=P(n-1)+2P(n-2)+2P(n-3)+...+2P(0) +2  but then the book says the sequence also satisfies the recurrence relation  P(n)=2P(n-1)+P(n-2). I am not able to understand this part.
28049
I am stuck in solving this math problem. I need to maximize  $$ (a_1\cdot x_1 + a_2\cdot x_2 + a_3\cdot x_3 + \dots + a_n\cdot x_n) \mathbin{\mathrm \%} b. $$ Here $a_i$ and $b$ are given and $x_i$ can be any nonnegative integer.
100982
I'd be interested in knowing if there are any problems that are easier to solve in a higher dimension, i.e. using solutions in a higher dimension that don't have an equally optimal counterpart in a lower dimension, particularly common (or uncommon) geometry and discrete math problems.
34010
$$T(n) = \sqrt{n} T \left(\sqrt n \right) + n$$  Master method does not apply here. Recursion tree goes a long way. Iteration method would be preferable.  The answer is $Θ (n \log \log n)$.  Can anyone arrive at the solution.
49456
Define $$L(w,u)=\frac{1}{2}\|w-u\|^2+\beta \left\|\frac{w}{x}\right\|,~w,u\in \Bbb{R}^n$$ where $$\frac{w}{x}=\left(\frac{w_1}{x_1},\ldots, \frac{w_n}{x_n}\right)$$ $$\|x\|=\sqrt{x_1^2+\cdots+x_n^2}$$ Given $u$, $x$ and $\beta$, how to get $$\arg\min_{w\in \Bbb{R}^n}L(w,u)$$
97529
Express the complex numbers given in Problem in the form $a+ib$ $i + i^2 + i^3 + \ldots + i^n \quad    (n &gt; 4)$
120059
how are we supposed to do this question if resistance is given as 1 ohm
32830
I have the following solution for my question but I am not sure if I have done the steps right:
59684
I have encountered this question in a coding competition and want the formula for $f(n)$. Is there any way to do this?
111691
I have been racking my brain to understand the solutions of constrained stars and bars problems like:  $a+b+c+d = 19$;  $a,b,c,d \in \{0,1,2,3,4,5,6,7\}$  Finding the total number of solutions is really simple $\dbinom{22}{3}$.   What to do after that? I want to understand the entire procedure step by step.   Please do not mark it as a duplicate of .  Reasons:   The accepted answer's trick doesn't always work.   Another person (the answer with two upvotes) has just listed out the binomial coefficients and I can't even remotely understand what he has tried to do.   The second answer isn't my cup of tea.
47039
Just doing this for revision, seems much harder than it should be, should I use $x=\tan u$ ? Any help appreciated.
37847
Actually I have no idea how to begin. I tried to represent it as a product of two series or as a sum of two series but have no result with this. Any idea ?
26817
\begin{multline*} \displaystyle \left \| f  \right \|_{M\dot{K}^{\alpha(\cdot),q}_{p(\cdot), \lambda}(\mathbb{R}^{n})} \approx \max \left\{ \sup_{L\leqslant0,L\in  \mathbb{Z}}2^{-L\lambda}\left ( \sum_{k=-\infty}^{L}2^{k\alpha(0)q}\left \|  f\chi_{k} \right \|^{q}_{L^{p(\cdot)}} \right )^{1/q},\\\sup_{L&gt;0,L\in \mathbb{Z}}\left[ 2^{-L\lambda}\left ( \sum_{k=-\infty}^{-1}2^{k\alpha(0)q} \left \| f\chi_{k} \right \|^{q}_{L^{p(\cdot)}} \right )^{1/q}\\ +2^{-L\lambda}\left ( \sum_{k=0}^{L}2^{k\alpha(\infty)q}\left \|  f\chi_{k} \right \|^{q}_{L^{p(\cdot)}} \right )^{1/q}\right] \right\}  \end{multline*}
80480
Like $\log(1+x)$? Is there any algorithms? I have read many materials but doesn't have an idea
837
Im trying to solve the following question         In the second step  where do they get $k!=2^k-1?$
13992
This is such a hard question to get my head around. Can anyone help solve this?
4462
For the bisection method, determine the formula $$n \geq \frac{\log(b_0-a_0)-\log(\epsilon)}{\log(2)}-1$$ for the number of steps to guarantee that $|{r-c_n}| &lt; \epsilon.$  This is one of the questions from my numerical analysis class. I understand how to use the bisection method, but I don't understand the analysis of the method or how to start deriving an answer.