# Problems from Definition and general term of a sequence

Verify whether the following numbers belong to every sequence, and in affirmative case state what position they occupy:

a) $$25$$. Does it belong to the sequence $$a_n=(3n+4)_{n\in\mathbb{N}}$$?

b) $$\dfrac{9}{5}$$. Does it belong to the sequence $$c_n=\dfrac{n^2}{n+1}$$?

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### 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: $$\left. \begin{array}{c} c_n=\dfrac{9}{5} \\ c_n=\dfrac{n^2}{n+1} \end{array} \right\} \Rightarrow \dfrac{9}{5}=\dfrac{n^2}{n+1} \Rightarrow 9(n+1)=5n^2$$$

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

### 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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Give the first five terms of the sequences with the general term:

a) $$a_n=(-1)^n(n+2)$$

b) $$b_n=\dfrac{(n+1)(n-1)}{2n^2}$$

c) $$c_n=3n^2-12$$

d) $$d_0=0$$, $$d_1=1$$ and $$d_{n+1}=d_n+d_{n-1}$$ (the sequence of Fibbonacci).

See development and solution

### Development:

a) $$a_1=(-1)^1(1+2)=-1\cdot3=-3$$$$$a_2=(-1)^2(2+2)=+1\cdot4=4$$$ $$a_3=(-1)^3(3+2)=-1\cdot5=-5$$$$$a_4=(-1)^4(4+2)=+1\cdot6=6$$$ $$a_5=(-1)^5(5+2)=-1\cdot7=-7$$$b) $$b_1=\dfrac{(1+1)(1-1)}{2\cdot1^2}=\dfrac{1\cdot0}{2}=0$$$ $$b_2=\dfrac{(2+1)(2-1)}{2\cdot2^2}=\dfrac{3\cdot1}{8}=\dfrac{3}{8}$$$$$b_3=\dfrac{(3+1)(3-1)}{2\cdot3^2}=\dfrac{4\cdot2}{18}=\dfrac{4}{9}$$$ $$b_4=\dfrac{(4+1)(4-1)}{2\cdot4^2}=\dfrac{5\cdot3}{32}=\dfrac{15}{32}$$$$$b_5=\dfrac{(5+1)(5-1)}{2\cdot5^2}=\dfrac{6\cdot4}{50}=\dfrac{12}{25}$$$

c) $$c_1=3\cdot1^2-12=3-12=-9$$$$$c_2=3\cdot2^2-12=12-12=0$$$ $$c_3=3\cdot3^2-12=27-12=15$$$$$c_4=3\cdot4^2-12=48-12=36$$$ $$c_5=3\cdot5^2-12=75-12=63$$$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) $$a_n=(-3,4,-5,6,-7,\ldots)$$

b) $$b_n=\Big(0,\dfrac{3}{8},\dfrac{4}{9},\dfrac{15}{32},\dfrac{12}{25},\ldots\Big)$$

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

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

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