Question

1. Write the first three terms of the following sequences whose nth terms are given by

Answer 1

HELLO DEAR,

( 1 ). $\bold{a_n = \frac{n(n - 2)}{3}}$
where, n = 1 , 2 , 3 ,......
Put n = 1
$\bold{a_1 = \frac{1(1 - 2)}{3}}$

$\bold{a_1 = \frac{-1}{3}}$
put n = 2
$\bold{a_2 = \frac{2(2 - 2)}{3}}$

$\bold{a_2 = 0}$

Put n = 3,
$\bold{a_3 = \frac{3(3 - 2)}{3}}$

$\bold{a_3 = 1}$

( 2 ). $\bold{C_n = (-1)^n (3)^{n + 2}}$

Put n = 1
$\bold{C_1 = (-1)^1(3)^{1 + 2}}$

$\bold{C_1 = (-1)(27) = -27}$

Put n = 2
$\bold{C_2 = (-1)^2(3)^{2 + 2}}$

$\bold{C_2 = (1)(81) = 81}$

Put n = 3,
$\bold{C_3 = (-1)^3(3)^{3 + 2}}$

$\bold{C_3 = (-1)(243) = -243}$

( 3 ).$\bold{Z_n = \frac{(-1)^nn(n + 2)}{4}}$

Put n = 1
$\bold{Z_1 = \frac{(-1)^1 1(1 + 2)}{4}}$

$\bold{Z_1 = \frac{(-1)(3)}{4}}$

$\bold{Z_1 = \frac{-3}{4}}$

Put n = 2
$\bold{Z_2 = \frac{(-1)^2 2(2 + 2)}{4}}$

$\bold{Z_2 = \frac{(1)(2)(4)}{4}}$

$\bold{Z_2 = 2}$

Put n = 3
$\bold{Z_3 = \frac{(-1)^3 3(3 + 2)}{4}}$

$\bold{Z_3 = \frac{(-1)(15)}{4}}$

$\bold{Z_3 = \frac{-15}{4}}$

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Solution :

i) an = [n(n-2)]/3

1 ) if n = 1 ,

a1 = [ 1(1-2)]/3

= -1/3

2 ) if n = 2 ,

a2 = [ 2(2-2)]/3 = 0

3) if n = 3 ,

a3 = [ 3(3-2)]/3 = 1

ii ) Cn = (-1)ⁿ 3^n+2 ,

1) if n = 1 ,

C1 = ( -1 )¹ 3^1+2

C1 = - 3³ = -27

2) if n = 2 ,

C2 = ( -1 )² × 3^2+2

c2 = 3⁴ = 81

3) if n = 3,

C3 = ( -1 )³ 3^3+2

= - 3^5 = 243

iii ) zn = [(-1)ⁿ n(n+2)]/4

1 ) if n = 1 ,

z1 = [ (-1)¹ × 1(1+2)]/4

z1 = -3/4 ,

2 ) if n = 2,

z2 = [ (-1)² × 2(2+2)]/4

z2 = 2 ,

3) if n = 3

z3 = [ (-1)³ × 3(3+2)]/4

z3 = -15/4

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