**Binomial theorem worksheet :**

**Here we are going to see some practice questions using ****binomial theorem.**

**Binomial expansion for (x + a) ^{n} is,**

nc_{0}x^{n}a^{0 }+ nc_{1}x^{n-1}a^{1 }+ nc_{2}x^{n-2}a^{2 }+ .........+ nc_{n}x^{n-n}a^{0}

(1) Find the expansion of (2x + 3y)^{5}

(2) Find the expansion of (3a + 5b)^{5}

(3) Find the expansion of (a - 2b)^{5}

(4) Find the expansion of (2*x* – 5*y*)^{7}

**Question 1 :**

Find the expansion of (2x + 3y)^{5}

**Solution :**

**Comparing the given question with (x + a) ^{n}**

**we get x = 2x, a = 3y and n = 5**

**Since we have power 5, we are going to have 6 terms in the expansion **

**(2x+3y) ^{5}** = 5c

**While calculating the values of 5c _{0}, 5c_{1} ,......... we can follow the given tricks**

**The value of nc _{0} and nc_{n} is 1.**

**The value of nc _{1} is n.**

**The value of 5c _{2} means, in the numerator we should write 5 as 5 **

5c_{0} = 1, 5c_{1} = 5, 5c_{2 }= (5**⋅**4)/(2**⋅**1) ==> 20/2 = 10

5c_{3 }= (5**⋅**4**⋅**3)/(3**⋅**2**⋅**1) ==> 60/6 = 10

5c_{4 }= (5**⋅**4**⋅**3**⋅**2)/(4**⋅**3**⋅**2**⋅**1) ==> 5, 5c_{5} = 1

= 32x^{5 }+ 5(16x^{4})(3y)^{1 }+ 10(8x^{3})(9y^{2 }) + 10(4x^{2})(27y^{3}) + 5(2x)(81y^{4}) + 1(243 y^{5})

= 32x^{5 }+240x^{4}y^{ }+ 720x^{3}y^{2} + 1080x^{2}y^{3} + 810xy^{4} + 243 y^{5}

**Question 2 :**

Find the expansion of (3a + 5b)^{5}

**Solution :**

**Comparing the given question with (x + a) ^{n}**

**we get x = 3a, a = 5b and n = 5**

(3a + 5b)^{5}

**Since we have power 5, we are going to have 6 terms in the expansion **

= 5c_{0}(3a)^{5}(5b)^{0 }+ 5c_{1}(3a)^{4}(5b)^{1 }+ 5c_{2}(3a)^{3}(5b)^{2 }+ 5c_{3}(3a)^{2}(5b)^{3 }+ 5c_{4}(3a)^{1}(5b)^{4}+ 5c_{5}(3a)^{0}(5b)^{5}

5c_{0} = 1, 5c_{1} = 5, 5c_{2 }= (5**⋅**4)/(2**⋅**1) ==> 20/2 = 10

5c_{3 }= (5**⋅**4**⋅**3)/(3**⋅**2**⋅**1) ==> 60/6 = 10

5c_{4 }= (5**⋅**4**⋅**3**⋅**2)/(4**⋅**3**⋅**2**⋅**1) ==> 5, 5c_{5} = 1

= 5c_{0}(3a)^{5}(5b)^{0 }+ 5c_{1}(3a)^{4}(5b)^{1 }+ 5c_{2}(3a)^{3}(5b)^{2 }+ 5c_{3}(3a)^{2}(5b)^{3 }+ 5c_{4}(3a)^{1}(5b)^{4}+ 5c_{5}(3a)^{0}(5b)^{5}

= 243a^{5} + 5(81a^{4})(5b) + 10 (27a^{3})(25b^{2}) + 10(9a^{2})(125b^{3}) + 5(3a) (625b^{4}) + 1 (3125b^{5})

= 243a^{5 }+ 2025a^{4}b + 6750a^{3}b^{2 }+ 11250a^{2}b^{3 }+ 9375ab^{4}+ 3125b^{5}

Let us see the solution of next problem on "Binomial theorem worksheet".

**Question 3 :**

Find the expansion of (a - 2b)^{5}

**Solution :**

**Comparing the given question with (x + a) ^{n}**

**we get x = a, a = -2b and n = 5**

(a - 2b)^{5}

**Since we have power 5, we are going to have 6 terms in the expansion **

= 5c_{0}(a)^{5}(-2b)^{0 }+ 5c_{1}(a)^{4}(-2b)^{1 }+ 5c_{2}(a)^{3}(-2b)^{2 }+ 5c_{3}(a)^{2}(-2b)^{3 }+ 5c_{4}(a)^{1}(-2b)^{4}+ 5c_{5}(a)^{0}(-2b)^{5}

= a^{5 }- 10 a^{4 }b^{ }+ 40 a^{3 }b^{2 }- 80 a^{2 }b^{3 }+ 80 a b^{4}-32 b^{5}

**Question 4 :**

Find the expansion of (2*x* – 5*y*)^{7}

**Solution :**

**Since we have power 7, we are going to have 8 terms in the expansion **

(2*x* – 5*y*)^{7}

= 7C_{0} (2*x*)^{7}(–5*y*)^{0} + 7C_{1} (2*x*)^{6}(–5*y*)^{1} + 7C_{2} (2*x*)^{5}(–5*y*)^{2}

+ 7C_{3} (2*x*)^{4}(–5*y*)^{3} + 7*C*_{4} (2*x*)^{3}(–5*y*)^{4} + 7*C*_{5} (2*x*)^{2}(–5*y*)^{5}

+ 7*C*_{6} (2*x*)^{1}(–5*y*)^{6} + 7*C*_{7} (2*x*)^{0}(–5*y*)^{7}

7c_{0} = 1, 7c_{1} = 7, 7c_{2 }= (7⋅6)/(2⋅1) ==> 56/2 = 28

7c_{3 }= (7⋅6⋅5)/(3⋅2⋅1) ==> 210/6 = 35

7c_{4 }= (7⋅6⋅5⋅4)/(4⋅3⋅2⋅1) = 35

7c_{5 }= (7⋅6⋅5⋅4⋅3)/(5⋅4⋅3⋅2⋅1) = 21

7c_{6 }= (7⋅6⋅5⋅4⋅3⋅2)/(6⋅5⋅4⋅3⋅2⋅1) = 7

7c_{7 }= 1

= (1)(128*x*^{7})(1) + (7) (64*x*^{6}) (–5*y*) + (21) (32*x*^{5}) (25*y*^{2}) + (35) (16*x*^{4}) (–125*y*^{3}) + (35) (8*x*^{3}) (625*y*^{4}) + (21) (4*x*^{2})(–3125*y*^{5}) + (7) (2*x*) (15625*y*^{6}) + (1)(1)(–78125*y*^{7})

= 128*x*^{7} – 2240*x*^{6}*y* + 16800*x*^{5}*y*^{2} – 70000*x*^{4}*y*^{3} + 175000*x*^{3}*y*^{4} – 262500*x*^{2}*y*^{5 }+ 218750*xy*^{6} – 78125*y*^{7}

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