Problems from Definition and general term of a sequence

Development:

a) The number $25$ will belong to the sequence if there exists a natural number $n$ in such a way that $a_n=25$. Since $a_n=3n+4$, replacing an equality in the other one we have: $$25=3n+4$$ $$3n=21$$ $$n=7$$

So, not only do we see that it belongs to the sequence, but we have figured out which position it occupies.

b) We will proceed as in the previous case: $$\begin{array}{c}{c}_n={\displaystyle \frac{9}{5}}\\ c_n={\displaystyle \frac{n^2}{n+1}}\end{array}\}\Rightarrow {\displaystyle \frac{9}{5}}={\displaystyle \frac{n^2}{n+1}}\Rightarrow 9(n+1)=5n^2$$

So we can only solve the equation of the second grade:

$$5n^2-9n-1=0\Rightarrow n={\displaystyle \frac{9\pm 3\sqrt{5}}{10}}$$ Since we have not obtaind any natural value for $n$, the number ${\displaystyle \frac{9}{5}}$ is not part of the sequence (it cannot occupy an irrational position!).

Solution:

a) $a_7=25$

b) It does not belong to the sequence.

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Development:

a) We relate every term to the position it occupies: $$a:\mathbb{N}\to \mathbb{R}$$ $$1\to 1$$ $$2\to 4$$ $$3\to 9$$ $$4\to 16$$ $$\dots$$

If we focus on the first term, we see that this one has not been modified, and we do not get any information. But if we focus on the second term, we see that $a_2$ is the double of $2$, $a_3$ is the triple of $3$, and $a_4$ is the result of $4\cdot 4$, that is, in order to obtain every term, we have to multiply the position of the term by itself, so we have raised it to the power of two: $$a_1=1^2=1,a_2=2^2=4,a_3=3^2=9,a_4=4^2=16,\dots$$ And as the general term we have: $$a_n=(n^2{)}_{n\in \mathbb{N}}$$

b) We will define this sequence in a recursive form. On the one hand, we observe that every term changes the sign in comparison to the previous one, and on the other hand, we can see that every term is double the previous one, so we have: $$a_{n+1}=-2\cdot a_n$$, and like that $a_1=-2$, we already have the sequence defined. (Also it is possible to give the general term $a_n=(-2{)}^n$ related to progressions.)

Solution:

a) $a_n=(n^2{)}_{n\in \mathbb{N}}$.

b) $a_{n+1}=-2\cdot a_n$, with $a_1=-2$.

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Development:

a) $$a_1=(-1{)}^1(1+2)=-1\cdot 3=-3$$ $$a_2=(-1{)}^2(2+2)=+1\cdot 4=4$$ $$a_3=(-1{)}^3(3+2)=-1\cdot 5=-5$$ $$a_4=(-1{)}^4(4+2)=+1\cdot 6=6$$ $$a_5=(-1{)}^5(5+2)=-1\cdot 7=-7$$

b) $$b_1={\displaystyle \frac{(1+1)(1-1)}{2\cdot 1^2}}={\displaystyle \frac{1\cdot 0}{2}}=0$$ $$b_2={\displaystyle \frac{(2+1)(2-1)}{2\cdot 2^2}}={\displaystyle \frac{3\cdot 1}{8}}={\displaystyle \frac{3}{8}}$$ $$b_3={\displaystyle \frac{(3+1)(3-1)}{2\cdot 3^2}}={\displaystyle \frac{4\cdot 2}{18}}={\displaystyle \frac{4}{9}}$$ $$b_4={\displaystyle \frac{(4+1)(4-1)}{2\cdot 4^2}}={\displaystyle \frac{5\cdot 3}{32}}={\displaystyle \frac{15}{32}}$$ $$b_5={\displaystyle \frac{(5+1)(5-1)}{2\cdot 5^2}}={\displaystyle \frac{6\cdot 4}{50}}={\displaystyle \frac{12}{25}}$$

c) $$c_1=3\cdot 1^2-12=3-12=-9$$ $$c_2=3\cdot 2^2-12=12-12=0$$ $$c_3=3\cdot 3^2-12=27-12=15$$ $$c_4=3\cdot 4^2-12=48-12=36$$ $$c_5=3\cdot 5^2-12=75-12=63$$

d) $$d_0=0$$ $$d_1=1$$ $$d_2=d_1+d_0=1+1=2$$ $$d_3=d_2+d_1=2+1=3$$ $$d_4=d_3+d_2=3+2=5$$ $$d_5=d_4+d_3=5+3=8$$ $$d_6=d_5+d_4=8+5=13$$

Solution:

a) $a_n=(-3,4,-5,6,-7,\dots )$

b) $b_n=(0,{\displaystyle \frac{3}{8}},{\displaystyle \frac{4}{9}},{\displaystyle \frac{15}{32}},{\displaystyle \frac{12}{25}},\dots )$

c) $c_n=(-9,0,15,36,63,\dots )$

d) $d_n=(0,1,2,3,5,\dots )$

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