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4

Hint: Check that $\displaystyle \forall n\geq 0, x_{n+1}-x_n=\frac{(-1)^n}{n+1}$ The series $\displaystyle \sum (x_{n+1}-x_n)$ is therefore convergent, and so is the sequence $(x_n)$, say $x_n\to l$ Furthermore, $\displaystyle l= \sum_{k=0}^\infty (x_{k+1}-x_k) =\sum_{k=0}^\infty \frac{(-1)^k}{k+1} = \log 2$. Edit: from a simulation on Mathematica, ...


4

Hint. From $$ a_{n+1} = \frac{1}{\frac{1}{a_n}+n} $$ you get $$ \frac1{a_n}-\frac1{a_{n+1}}=-n $$ then use a telescoping sum and a standard sum.


2

Let $b_n=1/a_n$ and find the recurrence satisfied by it.


2

I just wish to contribute a "quicker" development of Ragib Zaman's derivation. Just as he had shown, $$ \cos^k \theta = \left( \frac{ e^{i\theta} + e^{-i\theta} }{2} \right)^k = \frac{1}{2^k} \sum_{n=0}^k \binom{k}{n} (e^{i\theta} )^n (e^{-i\theta})^{k-n} = \frac{1}{2^k } \sum_{n=0}^k \binom{k}{n} e^{i(2n-k)\theta} $$ Now, assuming that $\theta$ is real, we ...


2

If the first two terms are $a$ and $b$ then you can add up terms until you get to the 7th. It will be: $$t_7=8a+13b$$ Similarly you can write down and add up the first 10 terms to get: $$s_{10}=\sum_{n=1}^{10}t_n=88a+143b$$ Solving the equality now gives: $$11(8a+13b)=88a+143b$$ As both sides are equal the first two numbers can be anything.


2

Hint: Let $a_n$ be the number of strings which satisfy the conditions of length $n$. If a string begins with $b$ or $c$, then you are left with a string of length $n-1$, which can start with any character. How many strings are there of this type (it is related to $a_{n-1}$)? If a string begins with $a$, then the second letter must be a $b$ or $c$, then ...


1

It's a second order differential equation, so there are $2$ arbitrary constants in the general solution. As you could see, there was a constraint on $c_2$, but there is none on $c_0$ and $c_1$, and there is no mutual constraint between the latter two constants. It seems most convenient to separate things out into $2$ series, one where all the exponents are ...


1

We can see that each solution is completed characterized by two numbers $c_0$ and $c_1$. Also, the terms that are influeced by $c_0$ do not overlap with the terms that are influenced by the terms influenced by $c_1$. In particular observe the relationship: $$y(c_0,c_1)=c_0y(1,0)+c_1y(0,1)$$


1

As a complement to the previous answers, observe that you can give initial conditions to select a unique solutions. The choice of initial data corresponds with the choice of the constanta $c_0$ and $c_1$: $c_0=1$ and $c_1=0$ corresponds to $y(0)=1$ and $y'(0)=0$. $c_0=0$ and $c_1=1$ corresponds to $y(0)=0$ and $y'(0)=1$.


1

The general theory about second order homogeneous differential equations says that (i) there is an open interval $I$ on the $x$-axis containing containing $0$ in its interior such that the given ODE has a two-dimensional vector space ${\cal L}$ of solutions $y(\cdot):\>I\to{\mathbb R}$, and (ii) these solutions are even real analytic, i.e. have convergent ...


1

If one denotes the first terms by $a_0$ and $a_1$, the $7$th term will be a linear combination of those, say $p_7 a_0+ q_7 a_1$. Same goes for the sum: $s_{10} =u_{10} a_0+ v_{10} a_1$. You do not really need to compute explicitely those linear terms. If you choose peculiar solutions: $a_0=0$ and $a_1=1$, the property works ($t^{0,1}_7=8$, ...


1

To gain intuition (and get the "right" solution), it is simpler to start consider the case where $n$ is a power of $2$, i.e. $n=2^k$ for some integer $k \geq 0$. (In what follows, all logarithms are in base $2$, because computer science.) We can write $$\begin{align} T(2^k) &= 16 T(2^{k-1}) + 2\cdot 2^{4k} = 2^4 T(2^{k-1}) + 2^{4k+1} \\ &= 2^4 ...


1

Suppose we seek to evaluate $$S(N) = \sum_{n=N}^\infty \sum_{k=N}^n \sum_{j=0}^k {k\choose j} (-1)^{k-j} (1+j)^n \frac{t^n}{n!}.$$ This is $$S(N) = \sum_{n=N}^\infty \frac{t^n}{n!} \sum_{k=N}^n \sum_{j=0}^k {k\choose j} (-1)^{k-j} (1+j)^n.$$ Now introduce $$(1+j)^n = \frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \exp((1+j)z) \; dz.$$ We ...


1

Let $x=A_\text{before};y=B_\text{before}$ $x'=A; y'=B$ $\lambda=C_{A\rightarrow B}; \mu=C_{B\rightarrow A}$ From equations given: $$\begin{align} \left. \begin{array}{1} x&=\dfrac{x'-(-\mu y)}{1-\lambda}\\ y&=\dfrac{y'-\lambda x}{1-\mu} \end{array} \right\}\\\\ \left. \begin{array}{1} (1-\lambda)x-\mu y&=x'\\ (1-\mu) y+\lambda ...


1

Hint: Use the *principle of superposition of solutions; find particular solutions of $$a_n=5a_{n-1} - 6_{n-2} + 4^n\quad (\text{resp. }{} + 2n, {}+ 3)$$ then add these particular solutions. Example for $4^n$: A particular solution will have the form $a_n=C 4^n$. It is indeed a solution if $$C4^n=5C4^{n-1}-6C4^{n-2}+4^n\iff 16C=20C-6C+16\iff C=8.$$


1

Note that the sequence satisfies the recurrence relation $x_{n+1} = x_n + \frac1{x_n} = \frac{x_n^2 + 1}{x_n}$ or $x_{n+1}x_n = x_n^2 + 1$. So if it converges, the limit $x$ must satisfy $$x^2 = x^2 + 1$$ which cannot be. Therefore the sequence cannot converge. Note that the sequence is always positive, so $1/x_n > 0$ and $x_{n+1} > x_n$, so the ...



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