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6

By setting $t_n=a_n+1$ we have: $$ a_{n+1} = a_n^2-2 \tag{1}$$ then by setting $a_n=2b_n$ we have: $$ b_{n+1} = 2 b_n^2-1 \tag{2} $$ It is not difficult to recognize the duplication formula for the (hyperbolic) cosine. In particular, assuming $b_0=\cosh t_0$, $b_n = \cosh(2^n t_0)$ follows by induction. Since in our case $b_1=\frac{3}{2}$ gives ...


4

You are almost there $$ a_{n+1} = a_1 + nd\\ a_n = a_1 + (n-1)d $$ thus we get $$ a_{n+1}a_n = \left(a_1 + nd\right)\left(a_1 + (n-1)d\right) $$ re-arrange your given relation you find $$ a_{n+1}a_n =9n^2-21n + 10 $$ can you equate coefficients?


3

If you have a recursion of the form $a_{n+1} =f(a_n) $, if $L = \lim_{n \to \infty} a_n $, then we must have $L = f(L)$. In your case, $f(x) = \sqrt{2+x}$. Therefore, for any limit $L$, we must have $L = \sqrt{2+L}$. Squaring, $L^2 = L+2$, which is a standard quadratic equation. Completing the square, from $L^2-L = 2$ we get $L^2-L+1/4 = 2+1/4 =9/4$, so ...


2

Hint: prove by induction that: $$t_n=\alpha^{2^n}+\alpha^{-2^n}+1 $$ with $\alpha$ is one of the roots of $x+\frac{1}{x}=3$


2

We have $$a_2=\frac{9-21+10}{a_1}\Rightarrow a_1a_2=-2\tag1$$ and $$a_3=\frac{36-42+10}{a_2}\Rightarrow a_2a_3=4\tag2$$ Since we have $a_1+a_3=2a_2$, with $(1)(2)$, we have $$a_1+\frac{4}{a_2}=2a_2\Rightarrow a_1a_2+4=2a_2^2\Rightarrow a_2=\pm1.$$ Can you take it from here?


2

It will be convenient to have all of the terms on the same side of the equation, so I will write your recurrence relation as $$ \tag{$\star$} a_n+\alpha_1a_{n-1}+\ldots+\alpha_k \,a_{n-k}=0, $$ and similarly for $b_n$. The characteristic polynomial of the recurrence $(\star)$ is $T^k+\alpha_1 T^{k-1}+\ldots+\alpha_k$. Every solution to $(\star)$ is a linear ...


2

There is no in my knowledge a name for these numbers. Using the formula you have given, the sums can be computed and we can conclude that : $$T(p,q)={p+q-1\choose q}+{q+p-1 \choose p}={p+q \choose p}$$ (when $p$ and $q$ are both positive)


2

Consider $t_{n+1} = (1+c~q^{n})~p~t_{n} + a + b + (n+1)bq$ for which \begin{align} \sum_{n=0}^{\infty} t_{n+1} \, x^{n} &= p \, T(x) + c \, p \, T(q x) + \frac{a + b}{1-x} + b q \, \partial_{x} \left( \sum_{n=0}^{\infty} x^{n} \right) \\ \frac{T(x) - t_{0}}{x} &= p \, T(x) + c \, p \, T(q x) + \frac{a + b}{1-x} + b q \, \partial_{x} \frac{1}{1-x} ...


1

You can divide both members by $3^n$ to get $\frac{M(n)}{3^n}=\frac{M(n-1)}{3^{n-1}}+\frac{1}{3^n}$. Let $N(n)=\frac{M(n)}{3^n}$. Thus we have $N(n)-N(n-1)=\frac{1}{3^n}$. It follows that $$N(n)-N(1)=\sum_{j=1}^{n-1}(N(j+1)-N(j))=\sum_{j=1}^{n-1}\frac{1}{3^{j+1}}.$$ So, $N(n)=\sum_{j=0}^{n-1}\frac{1}{3^{j+1}}$ and $M(n)=\frac{3^n-1}{2}.$


1

taking two consecutive terms: $$ M(n)=3M(n−1)+1 \\ M(n+1)=3M(n)+1 $$ and subtracting: $$ M(n+1) -M(n) = 3(M(n)-M(n-1)) $$ so $$ M(n+1) - M(n) = 3^{n-1}(M(2)-M(1)) = 3^n $$ similarly: $$ M(n)-M(n-1) = 3^{n-1} $$ adding all such equations: $$ M(n+1) -M(1)= \sum_{k=1}^n 3^k $$ i.e. $$ M(n+1) = \sum_{k=0}^n 3^k = \frac{3^n -1}2 $$


1

For large enough $n$, $$ M(n)=3M(n-1)+1\\ M(n)=3(3M(n-2)+1)+1\\ M(n)=9M(n-2)+4\\ M(n)=9(3M(n-3)+1)+4\\ M(n)=27M(n-3)+13\\ M(n)=27(3M(n-4)+1)+13\\ M(n)=81M(n-4)+40\\ M(n)=81(3M(n-5)+1)+40\\ M(n)=243M(n-5)+121\\ $$ Keep going until you notice a pattern. Eventually you will realize that $M(n)=3^{n-1}M(1)+\frac{3^{n-1}-1}{2}=\frac{3^n-1}{2}$. You can prove ...


1

The coefficients are given here. See the section on the formula: a(n,m)=2^(m-1)*n*(-1)^[(n-m)/2]*[(n+m)/2-1]!/{[(n-m)/2]! m!} if n>0. - R. J. Mathar, Apr 20 2007


1

$$ a_{n+1}=\frac{9n^2-21n+10}{a_n}\\a_{n+1}a_{n}=9n^2-21n+10\\(a_1+nd)(a_1+(n-1)d)=9n^2-21n+10\\a_1^2+(2n-1)a_1d+(n^2-n)d^2=9n^2-21n+10\\n^2d^2+n(-d^2+2a_1d)+(a_1^2-a_1d)=9n^2-21n+10$$ and now $$\left\{\begin{matrix} d^2=9 & \\ -d^2+2a_1d=-21& \\ a_1^2-a_1d=10& \end{matrix}\right. $$ you can find $d=\pm 3$ first ,and put d to find $a_1$


1

Notice that $$\frac{1-r^n}{1-r} = \sum_{k=0}^{n-1} r^k$$ Then it's easy to see that : $$T(1) = rT(0)+a$$ $$T(2) = r(rT(0)+a)+a = r^2T(0) + a(1 + r)$$ $$T(3) = r(r^2T(0) + a(1 + r)) + a = r^3T(0) + a(1 + r + r^2)$$ So you get by induction $$T(n) = r^nT(0) + a(1 + r + r^2 + \cdots + r^{n-1})$$ $$T(n) = r^nT(0) + a\frac{1-r^n}{1-r}$$


1

I would have set the relation as $$C_n = 2a C_{n-1} - C_{n-2}$$ with conditions $C_1=a$, $C_2=b$. The characteristic equation being $r^2=2ar-1$, the roots of it are given by $$r_{1,2}=a\pm\sqrt{a^2-1}$$ and so $$C_n=c_1 r_1^n+c_2 r_2^n$$ Applying the conditions $C_1=a$, $C_2=b$ then gives $$c_1=\frac{a r_2-b}{r_1 (r_2-r_1)}$$ $$c_2=\frac{b-a r_1}{r_2 ...


1

$$L=\sqrt{2+L}$$, squaring both side we have, $$L^2-L-2=0$$ $$(L-2)(L+1)=0$$ which gives $L=2,-1$ since $a_n\gt 0$, for all $n\in \Bbb N$, Hence $L=2$


1

We start from the recurrence $$T_n = c_1 + T_{n-1} + 2\sqrt{c_2 + c_1T_{n-1}}.$$ Multiplying that with $c_1$ - that is legitimate since $c_1 > 0$ - and adding $c_2$ yields \begin{align} c_2 + c_1 T_n &= c_2 + c_1^2 + c_1 T_{n-1} + 2c_1 \sqrt{c_2 + c_1 T_{n-1}}\\ &= c_1^2 + 2c_1\sqrt{c_2 + c_1 T_{n-1}} + (c_2 + c_1 T_{n-1})\\ &= (c_1 + ...


1

Since $$\sum\limits_{i=1}^{k} x_i =x_1 +x_2+\cdots + x_k$$ Then $$\sum\limits_{i=1}^{\color{red}{k+1}} x_i =x_1 +x_2+\cdots + x_k+\color{red}{x_{k+1}}$$ Therefore $$\sum\limits_{i=1}^{\color{red}{k+1}} x_i = \left(\sum\limits_{i=1}^{k} x_i\right)+\color{red}{x_{k+1}}$$



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