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1. ASSIGNMENT No. 2
Q. 1 (a)Reduce the matrix into echelon form
 1  3 4 3
 
 2  5 6 6
 
3 3 4 6
Course: M0
a
3
t
3
h
6-
e
4
m
646
a
7
t
3
i9
cs–I (1307)
Semester: Autumn, 2022
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3. 0314-4646739
(b)
x  y  z  8
Solve system 2x  3y  z  2
3x  2y  9z  9
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5. Q. 2
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(a) Find first four terms of the sequence:
4(n2
 2)
an   2
 n  2

 4
if n  3 or n > 6
if 3< n  6
Let 4(n2 − 2)
n = 3
4(32 − 2) = 28
n = 4
4(42 − 2) = 56
n = 5
4(52 − 2) = 102
n = 6
4(62 − 2) = 156
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6. 03
214-4646739
Let n −2
4
n = 3
32 − 2
4
n = 4
42 − 2
= 1.8
4
n = 5
52 − 2
= 3.5
4
n = 6
62 − 2
= 5.8
4
= 8.5
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So, S
th
ke
illin
se
g.
q
pu
kence is 1.8, 3.5, 5.8, 8.5, 28, 56, 1D
0i2
ya
,.1
pk
56
6
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7. (b) 03W
14-h
46
a4
t6
t7
e3
r
9
m of the sequence 2(n  2)
an 
n  4n  0
4
h
33
a6
s-4
t6
h
4
e
67
v3
a
9
lue ‘41’.
2
n2 + 4n + 4
2n + 4
= 41
n2 + 4n + 4 = 41 (2n + 4)
n2 + 4n + 4 − 41 (2n + 4) = 0
n2 + 4n + 4 − 82n − 164 = 0
n2 − 78n − 160 = 0
Replace n with x
x2 − 78x − 160 = 0
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9. x = n = 80 and -2. So, n cannot be negative so n = 80 is the term of sequence which have answer
41.
Q. 3 (a)The commissioners of 5 divisions meet to discuss the law-and-order situation. In how
many ways can they be seated at a round table, when three commissioners insist to sit
together?
Suppose The 5 persons are A, B, C, D, E
Total number of ways that 5 persons can sit at a round table (m)=
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10. The0
re
3’
1
s
4-
1
46
w
4a
6y
739
to sit A because all seats lo0
o3
k36
a
-l
4
i6
k4
e67
a3
t9an empty table. Then, ther0
e33
a2
r-
e
46
24w
67
a
3
y
9s to
seat B, 3 ways to seat C, 4 ways to seat D, and 2 ways to seat E, say total ways so that two of the people
do not sit together (n)=1×2×3×4×2=48
Hence, the resultant =m−n=120−48=72.
In 72 ways, 3 persons can sit at a round table, if two of the people do not sit together.
(b) Determine the probability of getting 2 Heads in three successive tosses of a balanced
coin.
Sample space: {HHH, HTH, THH, TTH, HHT, HTT, THT, TTT}
Total number of possible outcomes=8
Probability of getting at least two heads is:
P(A)=P (getting two heads) + P (getting 3 heads)
= 3/8 + 1/8
= 4/8 = 1/2
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11. 3 5
x2
p : x1
p 7 : 5
Missing variable P3 and P5 in this question so solution not possible.
(b) Show that whether x2 + y2 in a factor of x2k + y2k ; x  y, k  Ni or not.
x2k + y2k
Let k = 1
x2 + y2
True for k = 1.
Let k = k
x2k + y2k = m (x2 + y2)…………….1
k = k + 1
L.H.S
x2(k+1) + y2(k+1)
x2k+2 + y2k+2
Q. 4
03(
1a
4)
-4
F
6
i4
n
6
d
73
v
9alue of ‘x’ when
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

12. x2k.0
x
3
2
1
+
4-y
42
6k
4
.6y
72
39
From 1
x2k= m (x2 + y2) - y2k …………….1
(m (x2 + y2) - y2k). + y2k. y2
x2 (m (x2 + y2)) - x2 y2k + y2k. y2
x2 (m (x2 + y2)) – y2k (x2 - y2)
P (k+1) is true whenever p(k) is true.
Q. 5 (a) Expand: p3
 p 35
 p2
 p3
 23
From binomial distribution
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14. =
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15. 2
(b) Find, (3n – 5) the term of the end in the expansion of q 
 

 1 
2
2q 
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