UPSEE Solved Model Paper 4

Apply districuting rule
(a+b)3 = a3+3a2b+3ab2+b3
a = k2 , b = k
= (k2)3+3(k2)2k+3k2k2+k3
simplify = k6+3k5+3k4+k3
=

∞

Σ

k=1

3k2+3k+1

k6+3k5+3k4+k3

for an infinite upper boundary , if
an → 0 , then

∞

Σ

n=k

(an+1−an) = - ak

3k2+3k−1

k6+3k5+3k4+k3

= - (

1

(k+1)3

)−(−

1

k3

)
an+1 = −(

1

(k+1)3

)
an = −

1

k3

ak = a1 = −

1

13

lim

k→∞

(−

1

k3

)

lim

x→a

[c.f(x)] = c .

lim

x→a

f (x)
= −

lim

k→∞

(

1

k3

)

lim

k→a

[

f(x)

g(x)

] =

lim x→a f(x)

lim x→a g(x)

,

lim

x→a

g (x) ≠ 0
With the exception of indeterminate form
= −

lim k→∞ (1)

lim k→∞ (k3)

lim

k→∞

(1)

lim

x→a

c = c
= 1

lim

k→∞

(k3)
[Applying infinity property]

lim

x→∞

(axn+...+bx+c) = ∞, a > 0 , n is odd
= 1 , n = 3
= ∞
= -

1

∞

= 0
By the telescoping series test :

∞

Σ

k=1

3k2+3k+1

k6+3k5+3k4+k3

= −ak
= −(−

1

13

)
= 1
Option D is correct answer