Principle of Management
Grade:11
Section: A/B/C/M/N
Worksheet Number:4
Subject:
Mathematics
Topic:
Name of teacher:
Ms. Sheeba Manoj
Date :26/6/13
REVISION WORKSHEET
Submission date:
3/9/13
Name of student:
PRINCIPLE OF MATHEMATICAL INDUCTION
Prove the following by Mathematical Induction
1.
1
1 + 4 + 7 + …………………….. + (3n – 2 ) = n(3n 1)
2
2.
4 + 8 + …………………………. + 4n = 2n( n + 1 )
3.
1
1.3 + 2.4 + 3.5 + ………………. + n(n +2) = n(n 1)(2n 7)
6
4.
5
5 + 15 + 45 +……………………... +5.3n – 1 = (3 n 1 )
2
5. n a + (a + d) + ( a + 2d) + ……………… + [a +(n – 1)d] = [2a (n 1)d ]
2
6.
1
12 + 32 + 52 +…………………………..+ (2n – 1)2 = n(4n 2 1)
3
7.
1
1
1
1 n
...................................
3.7 7.11 11.15
(4n 1)(4n 3) 3(4n 3) n + 15n – 1 is divisible by 9
8.
4
9.
10.
11.
32n + 7 is divisible by 8
2.7n + 3.5n – 5 is divisible by 24.
12.
13.
14.
7 n 3n is divisible by 4 n(n+1)(2n+1) is divisible by 6
15.
2n + 7 < ( n + 3 )2
x 2 n1 y 2 n1 is divisible by x + y.
1 + 2 + 3 + ……………………..+ n <
Sheeba Manoj
1
(2n 1) 2
8
Page 1
TRIGONONOMETRY
Sheeba Manoj
Page 2
COMPLEX NUMBERS
Prove that
1.
1 + i2 + i4 +i6 = 0
2.
6i54 + 5i37 - 2i11 + 6i68 = 7i
1
( 1+ i )4 ( 1 + )4 = 16 i 3.
4.
in. i n+1. i n+2 . i n+3 = -1 , n N
Solve for x and y.
5.
2x + 3iy = 5 + 6i
6.
( 3 + i ) x + ( 1 – 2i) y + 7i = 0
7.
1 i x 2i
3i
+
2 3i y i
3i
= i
Express the following numbers in the form x + iy
8.
2 3i
9.
1 i
2
1 i
3
10.
11.
1 i3
2
3
2
+
1 i 2 i 1 i
5 12i 5 12i
5 12i 5 12i
Find the multiplicative inverse of the following
12.
(2 – i) (3+i)
13.
2 3i
3 2i
14.
i 1i 2
i 1i 2
Sheeba Manoj
Page 3
Find the modulus and amplitude of the following complex numbers and express in the polar form.
3i
15.
1+
17.
Section: A/B/C/M/N
Worksheet Number:4
Subject:
Mathematics
Topic:
Name of teacher:
Ms. Sheeba Manoj
Date :26/6/13
REVISION WORKSHEET
Submission date:
3/9/13
Name of student:
PRINCIPLE OF MATHEMATICAL INDUCTION
Prove the following by Mathematical Induction
1.
1
1 + 4 + 7 + …………………….. + (3n – 2 ) = n(3n 1)
2
2.
4 + 8 + …………………………. + 4n = 2n( n + 1 )
3.
1
1.3 + 2.4 + 3.5 + ………………. + n(n +2) = n(n 1)(2n 7)
6
4.
5
5 + 15 + 45 +……………………... +5.3n – 1 = (3 n 1 )
2
5. n a + (a + d) + ( a + 2d) + ……………… + [a +(n – 1)d] = [2a (n 1)d ]
2
6.
1
12 + 32 + 52 +…………………………..+ (2n – 1)2 = n(4n 2 1)
3
7.
1
1
1
1 n
...................................
3.7 7.11 11.15
(4n 1)(4n 3) 3(4n 3) n + 15n – 1 is divisible by 9
8.
4
9.
10.
11.
32n + 7 is divisible by 8
2.7n + 3.5n – 5 is divisible by 24.
12.
13.
14.
7 n 3n is divisible by 4 n(n+1)(2n+1) is divisible by 6
15.
2n + 7 < ( n + 3 )2
x 2 n1 y 2 n1 is divisible by x + y.
1 + 2 + 3 + ……………………..+ n <
Sheeba Manoj
1
(2n 1) 2
8
Page 1
TRIGONONOMETRY
Sheeba Manoj
Page 2
COMPLEX NUMBERS
Prove that
1.
1 + i2 + i4 +i6 = 0
2.
6i54 + 5i37 - 2i11 + 6i68 = 7i
1
( 1+ i )4 ( 1 + )4 = 16 i 3.
4.
in. i n+1. i n+2 . i n+3 = -1 , n N
Solve for x and y.
5.
2x + 3iy = 5 + 6i
6.
( 3 + i ) x + ( 1 – 2i) y + 7i = 0
7.
1 i x 2i
3i
+
2 3i y i
3i
= i
Express the following numbers in the form x + iy
8.
2 3i
9.
1 i
2
1 i
3
10.
11.
1 i3
2
3
2
+
1 i 2 i 1 i
5 12i 5 12i
5 12i 5 12i
Find the multiplicative inverse of the following
12.
(2 – i) (3+i)
13.
2 3i
3 2i
14.
i 1i 2
i 1i 2
Sheeba Manoj
Page 3
Find the modulus and amplitude of the following complex numbers and express in the polar form.
3i
15.
1+
17.