VITEEE 2010 Paper

(3−5x)11=311(1−

5x

3

)11
=311(1−

5

3

.

1

5

)11 (∵‌x=

1

5

)
=311(1−

1

3

)11
Now, r=

|x|(n+1)

|x|+1

=

|− 1 3 |(11+1)

|− 1 3 |+1

=

4 4

3

⇒ r = 3
Therefore, 3rd (T3) and (3+1) = 4th (T4) terms are numerically greatest in the expansion of (3−5x)11.
So, greatest term = T3
=311|11C2(1)9(−

1

3

)2|
=311|

11×10

1.2.9

|
=55×39
and
T4=311|11C3(1)8(−

1

3

)3|
=311|

11×10×9

1.2.3

.(−

1

27

)|
=55×39
∴ Greatest term(numerically) =T3=T4=55×39