====== Question 1 and 2 Exercise 4.1 ======
Solutions of Question 1 and 2 of Exercise 4.1 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

=====Question 1(i)=====
Classify into finite and infinite sequences\\ $2,4,6,8, \ldots ,50$
====Solution==== 
It is finite sequence whose last term is $50 $.

=====Question 1(ii)=====
Classify into finite and infinite sequences.
$1,0,1,0,1, \ldots$.
====Solution====
It is infinite sequence, the last term may be $0$ or $1$ , but how much terms in this sequence, we don't know.

=====Question 1(iii)=====
Classify into finite and infinite sequences:
$...,-4,0,4,8, \ldots, 60$
====Solution====
This is infinite sequence.

=====Question 1(iv)=====
Classify into finite and infinite sequences.
$1,-\dfrac{1}{3}, \dfrac{1}{9},-\dfrac{1}{27}, \ldots,-\dfrac{1}{2187}$
====Solution====
This finite sequence.

=====Question 2(i)====
Find first four terms of the sequence with the given general terms:
$a_n=\dfrac{n(n+1)}{2}$
====Solution====
Given:
$$a_n=\dfrac{n(n+1)}{2}$$
For first term, put $n=1,$
$$a_1=\dfrac{1(1+1)}{2}=1$$
For second term, put $n=2$.
$$a_2=\dfrac{2(2+1)}{2}=3$$
For third term, put $n=3,$
$$a_3=\dfrac{3(3+1)}{2}=6$$
For forth term, put $n=4,$
$$a_4=\dfrac{4(4+1)}{2}=10$$
Hence first four terms of the sequence are $1,3,6, 10$.

=====Question 2(ii)====
Find first four terms of the sequence with the given general terms:
$a_n=(-1)^{n-1} 2^{n+1}$
====Solution====
Given: 
$$a_n=(-1)^{n-1} 2^{n+1}$$
For first term, put $n=1$, then
$$a_1=(-1)^{1-1} 2^{1+1}=2^2=4$$
For second term, put $n=2$, then
$$a_2=(-1)^{2-1} 2^{2+1}=-2^3=-8$$
For third term, put $n=3$, then
$$a_3=(-1)^{3-1} 2^{3-1}=2^4=16$$
For fourth term, put $n=4$, then
$$a_4=(-1)^{4 \cdots 1} 2^{4+1}=-2^5=-32 \text {. }$$
Hence first four terms of the sequence are $4,-8,16,-32$.

=====Question 2(iii)====
Find first four terms of the sequence with the given general terms:
$a_n =(\dfrac{1}{3})^n$
====Solution====
Given:
$$a_n =(\dfrac{1}{3})^n$$
For first term, put $n=1$,
$$a_1=(\dfrac{1}{3})^1=\dfrac{1}{3}$$
For second term, put $n=2$,
$$a_2=(\dfrac{1}{3})^2=\dfrac{1}{9}$$
For third term, put $n=3$,
$$a_3=(\dfrac{1}{3})^3=\dfrac{1}{27}$$
For fourth term, put $n=4$,
$$a_4=(\dfrac{1}{3})^4=\dfrac{1}{81}$$
Hence first four terms of the sequence are $\dfrac{1}{3},\dfrac{1}{9},\dfrac{1}{27},\dfrac{1}{81}$.
=====Question 2(iv)====
Find first four terms of the sequence with the given general terms:
$a_n=\dfrac{n(n-1)(n-2)}{6}$
====Solution====
Given:
$$a_n=\dfrac{n(n-1)(n-2)}{6}$$
For first term, put $n=1,$ $$a_1=0$$
For second term, put $n=2$, $$a_2=0$$
For third term, put $n=3$, $$a_3=1$$
For fourth term, put $n=4$, $$a_4=4$$
Hence first four terms of the sequence are $0,0,1,4$.




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