Question 5 and 6, Exercise 4.1

Solutions of Question 5 and 6 of Exercise 4.1 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

The $n$th term of the sequence is given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term: $a_{15}$.$a_{n}=n^{2}-2 n$

Solution. Given $$a_n = n^2 - 2n.$$ Then $$\begin{align*} a_1 &= (1)^2 - 2(1) = 1 - 2 = -1\\ a_2 &= (2)^2 - 2(2) = 4 - 4 = 0\\ a_3 &= (3)^2 - 2(3) = 9 - 6 = 3\\ a_4 &= (4)^2 - 2(4) = 16 - 8 = 8\\ \end{align*}$$ Now $$\begin{align*} a_{10} &= (10)^2 - 2(10) = 100 - 20 = 80\\ a_{15} &= (15)^2 - 2(15) = 225 - 30 = 195 \end{align*}$$ So, $a_1 = -1$,$a_2 = 0$, $a_3 = 3$, $a_4 = 8$, $a_{10} = 80$, $a_{15} = 195$.

The $n$th term of the sequence is given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term: $a_{n}=\frac{n^{2}-1}{n^{2}+1}$

Solution. Given $$a_n = \frac{n^2 - 1}{n^2 + 1}.$$ Then $$\begin{align*} a_1 &= \frac{1^2 - 1}{1^2 + 1} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0\\ a_2 &= \frac{2^2 - 1}{2^2 + 1} = \frac{4 - 1}{4 + 1} = \frac{3}{5}\\ a_3 &= \frac{3^2 - 1}{3^2 + 1} = \frac{9 - 1}{9 + 1} = \frac{8}{10} = \frac{4}{5}\\ a_4 &= \frac{4^2 - 1}{4^2 + 1} = \frac{16 - 1}{16 + 1} = \frac{15}{17}\\ \end{align*}$$ Now $$\begin{align*} a_{10} &= \frac{10^2 - 1}{10^2 + 1} = \frac{100 - 1}{100 + 1} = \frac{99}{101}\\ a_{15} &= \frac{15^2 - 1}{15^2 + 1} = \frac{225 - 1}{225 + 1} = \frac{224}{226} = \frac{112}{113} \end{align*}$$ So, $a_1 = 0$, $a_2 = \frac{3}{5}$, $a_3 = \frac{4}{5}$, $a_4 = \frac{15}{17}$, $a_{10} = \frac{99}{101}$, $a_{15} = \frac{112}{113}$.

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