**Question 1.Expand:(i) (√3 + √2)**

^{4}Solution:

Here, a = √3, b = √2 and n = 4.

Using binomial theorem,

∴ (√3 + √2)^{4} = 1(9) (1) + 4(3√3) (√2) + 6(3)(2) + 4(√3) (2√2) + 1(1)(4)

= 9 + 12√6 + 36 + 8√6 + 4

= 49 + 20√6

**(ii) (√5 – √2) ^{5}Solution:**

Here, a = √5, b = √2 and n = 5.

Using binomial theorem,

**Question 2.Expand:(i) (2x**

^{2}+ 3)

^{4}Solution:

Here, a = 2x

^{2}, b = 3 and n = 4.

Using binomial theorem,

Using binomial theorem,

**Question 3.Find the value of(i) (√3 + 1)**

^{4}– (√3 – 1)

^{4}Solution:

**(ii) (2 + √5) ^{5} + (2 – √5)^{5}**

Solution:

Adding (i) and (ii), we get

∴ (2 + √5 )

^{5}+ (2 – √5)

^{5}= (32 + 80√5 + 400 + 200√5 + 250 + 25√5) + (32 – 80√5 + 400 – 200√5+ 250 – 25√5 )

= 64 + 800 + 500

= 1364

**Question 4.Prove that:(i) (√3 + √2)**

^{6}+ (√3 – √2)

^{6}= 970Solution:

**(ii) (√5 + 1) ^{5} – (√5 – 1)^{5} = 352**

Solution:

**Question 5.Using binomial theorem, find the value of(i) (102)**

^{4}Solution:

**(ii) (1.1) ^{5}Solution:**

**Question 6.Using binomial theorem, find the value of(i) (9.9)**

^{3}Solution:

**(ii) (0.9) ^{4}Solution:**

**Question 7.Without expanding, find the value ofSolution:**

**(ii) (2x – 1) ^{4} + 4(2x – 1)^{3} (3 – 2x) + 6(2x – 1)^{2} (3 – 2x)^{2} + 4(2x – 1)^{1} (3 – 2x)^{3} + (3 – 2x)^{4}**

Solution:

**Question 8.Find the value of (1.02) ^{6}, correct upto four places of decimals.Solution:**

**Question 9.Find the value of (1.01) ^{5}, correct upto three places of decimals.Solution:**

**Question 10.Find the value of (0.9) ^{6}, correct upto four places of decimals.Solution:**