**Formulas:**

** **

1. The **general term** in the expansion of **(a + b) ^{n }:**

** T _{r + 1} = ^{n}C_{r }**

**×**

**(a)**

^{n – r}**×**

**b**

^{r}** **

2. The **middle term** in the expansion of **(a + b) ^{n}** :

** (a). If n is even:**

** The middle term = (n2+1)thterm**

** **

** (b). If n is odd:**

**The middle term = (n+12)thtermand(n+12+1)thterm **

** **

**Q.1: Find the Coefficient of x ^{6 }**in

**the expansion of (x + 2)**

^{9}

**Sol.**

**Since, ****The general term in the expansion of (a + b) ^{n }is given by: T_{r + 1} = ^{n}C_{r }**

**×**

**(a)**

^{n – r}**×**

**b**

^{r}Therefore, on substituting **n = 9, a = x and b = 2** in the above expression we will get:

**T _{r + 1} = **

^{9}**C**

_{r }**×**

**(x)**

^{9 – r}**×**

**2**

^{r}. . . . . . . (1)Now, on comparing the coefficient of x in equation (1) with x^{6}, we will get:

i.e. (x)^{9 – r} = x^{6 }

(or) 9 – r = 6

Therefore, **r = 3**

Now, on substituting the value of r in equation (1) we will get:

T_{4} = ^{9}C_{3 }× (x)^{9 – 3} × 2^{3
}

**Therefore, the Coefficient of x ^{6 }in the**

**expansion of (x + 2)**

^{9}= 672

**Q.2: Find the Coefficient of a ^{6 }b^{8 }in the**

**expansion of (2a – 3b)**

^{14}

**Sol.**

**Since, ****The general term in the expansion of (x + y) ^{n }is given by: T_{r + 1} = ^{n}C_{r }**

**×**

**(x)**

^{n – r}**×**

**y**

^{r}Therefore, on substituting **n = 14, x = 2a and y = (-3b) **in the above expression we will get:

T_{r + 1} = ^{14}C_{r }× (2a)^{14 – r} × (-3b)^{r}

i.e. **T _{r + 1} = [**

^{14}**C**

_{r }**×**

**(2)**

^{14 – r}(-3)^{r}] (a)^{14 – r}**×**

**(b)**

^{r}. . . . . . . . . . (1)Now, on comparing the **coefficients of a and b in equation (1) with a ^{6 }b^{8}**, we will get:

i.e. a^{6} b^{8} = a^{14 – r} b^{r}

Therefore, **r = 8**

Now, on substituting the value of **‘r’ in equation (1)** we will get:

T_{9} = [ ^{14}C_{8 }× (2)^{14 – 8} (-3)^{8} ] (a)^{14 – 8} × (b)^{8}

i.e. T_{9} = [ ^{14}C_{8 }× (2)^{6} (-3)^{8} ] a^{6} b^{8}

**Therefore, the Coefficient of a ^{6 }b^{8 }in the**

**expansion of (2a – 3b)**

^{14 }= 140107968** **

**Q.3: Write the general term in the expansion of (x ^{3} – y^{2})^{5}**

**Sol.**

The **general term** in the expansion of **(a + b) ^{n }**is given by

**: T**

_{r + 1}=^{n}C_{r }**×**

**(a)**

^{n – r}**×**

**b**

^{r}Now, on substituting **n = 5, a = x ^{3} and b = y^{2}** we will get:

**T _{r+1} = ^{5}C_{r }**

**×**

**(x**

^{3})^{5 – r}**× (**

**y**

^{2})^{r}**Therefore, the general term in the expansion of (x ^{3} – y^{2})^{5}:**

**T _{r+1} = ^{5}C_{r }**

**×**

**(x)**

^{15 – 3 r}**× (**

**y)**

^{2 r}** **

**Q.4: Write the general term in the expansion of (x ^{4} – xy^{2})^{9}**

**Sol.**

The **general term** in the expansion of **(a + b) ^{n }**is given by

**: T**

_{r + 1}=^{n}C_{r }**×**

**(a)**

^{n – r}**×**

**b**

^{r}Now, on substituting **n = 9, a = x ^{4} and b = xy^{2}** we will get:

**T _{r+1} = ^{9}C_{r }**

**×**

**(x**

^{4})^{9 – r}**× (x**

**y**

^{2})^{r}_{r+1} = ^{9}C_{r }× (x)^{36 – 4 r} × (x)^{2 r} × (y)^{2 r}

_{r+1} = ^{9}C_{r }× (x)^{36 – 4 r + 2 r} × (y)^{2 r}

_{r+1} = ^{9}C_{r }× (x)^{36 – 2 r }× (y)^{2 r}

**Therefore, the general term in the ****expansion of (x ^{4} – xy^{2})^{9}**

**:**

**T _{r+1} = ^{9}C_{r }**

**×**

**(x)**

^{36 – 2 r}**× (**

**y)**

^{2r}

**Q.5: Find the 4 ^{th} term in the expansion of (2x + 3y)^{10 }**

**Sol.**

**Since, **The **general term** in the expansion of **(a + b) ^{n }**is given by

**: T**

_{r + 1}=^{n}C_{r }**×**

**(a)**

^{n – r}**×**

**b**

^{r}Therefore, from the above equation the **value of r should be 3;** for finding out the values of 4^{th} Term (T_{4})

Therefore, **T _{3 + 1} = ^{n}C_{3 }**

**×**

**(a)**

^{n – 3}**×**

**b**

^{3}Now, on substituting **n =10, a = 2x and b = 3y** we will get:

_{3 + 1} = ^{10}C_{3 }× (2x)^{10 – 3} × (3y)^{3}

_{4} = ^{10}C_{3 }× (2x)^{7} × (3y)^{3}

**Therefore, 4 ^{th} term in the expansion of (2x + 3y)^{10} = 414720 x^{7}y^{3}**

** **

**Q.6 Find the 11 ^{th} term in the expansion of (8x−12x√)15 **

^{ }

**Sol.**

**Since, **The **general term** in the expansion of **(a + b) ^{n }**is given by

**: T**

_{r + 1}=^{n}C_{r }**×**

**(a)**

^{n – r}**×**

**b**

^{r}Therefore, from the above equation the **value of r should be 10** for finding out the values of 11^{th} Term (T_{11})

Therefore, **T _{10 + 1} = ^{n}C_{10 }**

**×**

**(a)**

^{n – 10}**×**

**b**

^{10}Now, on substituting **n =15, a = 8x and b = **** (−12x√) **we will get:

_{10 + 1} = ^{15}C_{10 }× (8x)^{15 – 10} ×

_{11} = ^{15}C_{10 }× (8x)^{5} ×

**Therefore, 11 ^{th} term in the expansion of (8x−12x√)15= 96096**

^{ }

**Q.7: In the expansion of (5−x410)9, Find the middle terms.**

**Sol.**

Here, **n = 9**

When **n** is **odd** then the **middle** **terms** in the expansion of **(a + b) ^{n}** are given by:

**middle terms** **=**

Therefore, the middle terms in the expansion of** (5−x410)9**

**are:**

^{th} term and 6^{th} term

**Now, 5 ^{th} and 6^{th} terms in the expansion of **

**are:**

**T _{5}** = T

_{4 + 1}=

^{9}C

_{4 }× (5)

^{9 – 4}×

_{5} = ^{9}C_{4 }× (5)^{5} ×

**Therefore, 5 ^{th} term =3158x16**

Now,** T _{6}** = T

_{5 + 1}=

^{9}C

_{5 }× (5)

^{9 – 5}×

_{6} = ^{9}C_{5 }× (5)^{4} ×

**Therefore, 6 ^{th} term =63160x20**

**Therefore, 5 ^{th} and 6^{th} terms are the middle terms in the expansion of **

(5−x410)9

**And also, 5 ^{th} term =3158x16 and 6^{th} term =−63160x20**

** **

**Q.8: In the expansion of (x2+8y)10, Find the middle term.**

**Sol.**

Here, **n = 10**

When **n is even**, the **middle term** in the expansion of **(a + b) ^{n}** is given by:

Therefore, the **middle term** in the expansion of** (x2+8y)10 is**:

^{th} term

**Now, 6 ^{th} term in the expansion of **

(x2+8y)10 is:

**T _{6}** = T

_{5 + 1}=

^{10}C

_{5 }×

^{5}

_{6} = ^{10}C_{5 }× ^{5}

**Therefore, 6 ^{th} term = 258048 x^{5 }y^{5}**

**Hence, ****the middle term in the expansion of (x2+8y)10 = 258048 x ^{5 }y^{5}**

** **

**Q.9: Prove that the coefficients of p ^{a }and p^{b} are equal in the expansion of (1 + p)^{a + b}**

**Sol.**

**Since, **The **general term** in the expansion of **(x + y) ^{n }**is given by

**: T**

_{r + 1}=^{n}C_{r }**×**

**(x)**

^{n – r}**× (**

**y)**

^{r}Now, on substituting **n = (a + b), x =1 and y = p,** we will get:

T_{r + 1} = ^{(a + b)}C_{r }× (1)^{(a + b) – r} × (p)^{r}

_{r + 1 }= ^{(a + b)}C_{r} **×**** (p) ^{r} . . . . . . . . . . . . (1)**

Now, on comparing the coefficient of p in equation (1) with p^{a}, we will get r = a

**Therefore, from equation (1):**

T_{ a + 1} = ^{(a + b)}C_{a} × (p)^{a}

**Therefore, the coefficient of (p) ^{a} = (a+b)!a!b! . . . . . . . . . . (2)**

Now, on comparing the coefficient of p in equation (1) with p^{b}, we will get r = b

Therefore, from equation (1):

T_{ b + 1} = ^{(a + b)}C_{b} × (p)^{b}

**Therefore, the coefficient of (p) ^{b} = (a+b)!b!a! . . . . . . . . . . (3)**

**Now, on comparing equation (2) and equation (3) we conclude that the coefficients of p ^{a }and p^{b} are equal in the expansion of (1 + p)^{a + b}.**

**Hence, Proved**

** **

**Q.10: The coefficients of the (k + 1) ^{th}, k^{th} and (k – 1)^{th} terms in the expansion of (x + 1)^{n} are in the ratio 5 : 3 : 1. Find the values of n and k.**

**Sol.**

**Since, ****The general term in the expansion of (a + b) ^{n }is given by: T_{r + 1} = ^{n}C_{r }**

**×**

**(a)**

^{n – r}**×**

**b**

^{r}Now on substituting **a = x and b = 1** in the above equation:

**T _{r + 1} = ^{n}C_{r }**

**×**

**(x)**

^{n – r}**× (**

**1)**

^{r}**Now, (k + 1) ^{th} term in the expansion of (x + 1)^{n} :**

T_{(k) + 1} = ^{n}C_{k }× (x)^{n – k} × 1^{(k )}

i.e. **T _{(k + 1)} = ^{n}C_{k }**

**×**

**(x)**

^{n – k}**Therefore the coefficient of T _{(k + 1)}^{th }term = **

**Now, (k) ^{th}**

**term in the expansion of (x + 1)**

^{n}:T_{(k – 1) + 1} = ^{n}C_{k – 1 }× (x)^{n – (k – 1)} × 1^{(k – 1)}

i.e. **T _{k} = ^{n}C_{k – 1} **

**×**

**(x)**

^{n – k + 1}**Therefore the coefficient of T _{(k)}^{th }term = **

**And, (k – 1) ^{th}**

**term in the expansion of (x + 1)**

^{n}:T_{(k – 2 )+ 1} = ^{n}C_{k – 2 }× (x)^{n – (k – 2)} × 1^{(k – 2)}

i.e. **T _{k – 1} = ^{n}C_{k – 2 }**

**×**

**(x)**

^{n – k + 2}**Therefore the coefficient of T _{(k – 1)}^{th }term = **

Now, according to the given conditions: **Coefficients of T _{(k + 1)}**,

**T**

_{k}and T_{k – 1}are in the ratio of**5 : 3 : 1**

**Therefore, Tk−1Tk=13**

**(or) ****n – 4k + 5 = 0 . . . . . . . . . . . . . . . . (1)**

And,** **

**(or) 3n – 8k + 3 = 0 . . . . . . . . . . . . . . . . . . (2)**

On **multiplying equation (1)** by **3** and **subtracting** it to **equation (2)** we will get:

(3n – 8k + 3) – 3(n – 4k + 5) = 0

**Therefore, k = 3**

Now on substituting the value of k in **equation (2)** we will get:

3n – 8(3) + 3 = 0

**Therefore, n = 7**

** **

**Q.11: Prove that the coefficient of a ^{p} in the expansion of (1 + 3a)^{2p – 1 }is half the coefficient of a^{p} in the expansion of (1 + 3a)^{2p}**

**Sol.**

**Since, **The **general term** in the expansion of **(x + y) ^{n }**is given by

**: T**

_{r + 1}=^{n}C_{r }**×**

**(x)**

^{n – r}**× (**

**y)**

^{r}**For (1 + 3a) ^{2p}:**

On putting n = 2p, x =1 and y = 3a, we will get:

T_{r + 1} = ^{(2p)}C_{r }× (1)^{(3p) – r} × (3a)^{r}

_{r + 1 }= ^{(2p)}C_{r} **×**** (3a) ^{r} . . . . . . . . . . . . (1)**

Now, on comparing the coefficient of a in equation (1) with a^{p}, we will get r = p

**Therefore, from equation (1):**

T_{ p + 1} = ^{(2p)}C_{p} × (3a)^{p}

**Therefore, the coefficient of a ^{p} =(2p)!p!p!×3p . . . . . . . . . . . . . . . . (2)**

Now, for **(1 + 3a) ^{2p – 1 }:**

On substituting n = (2p – 1), x =1 and y = 3a, we will get:

T_{r + 1} = ^{(2p – 1)}C_{r }× (1)^{(2p – 1) – r} × (3a)^{r}

_{r + 1 }= ^{(2p – 1)}C_{r} **×**** (3a) ^{r} . . . . . . . . . . . . (1)**

Now, on comparing the coefficient of ‘a’ in equation (1) with ‘a^{p }’, we will get **r = p**

Therefore, **from equation (1):**

T_{ p + 1} = ^{(2p – 1)}C_{p} × (3a)^{p}

**Therefore, the coefficient of a ^{p} = =(2p)!2(p!)2×3p . . . . . . . . . . (3)**

Now, on **comparing** **equation (2) and equation (3):**

^{(2p)}C_{p} **× ****3 ^{p }= 12 ^{(2p – 1)}C_{p} **

**×**

**3**

^{p}**Therefore, the coefficient of a ^{p} in the expansion of (1 + 3a)^{2p – 1} is half the coefficient of a^{p} in the expansion of (1 + 3a)^{2p}**

** **

**Q.12: For what values of ‘m’ the coefficient of x ^{2} in the expansion of (1+ 2x)^{k} is 140.**

**Sol.**

**Since, **The **general term** in the expansion of **(a + b) ^{n }**is given by

**: T**

_{r + 1}=^{n}C_{r }**×**

**(a)**

^{n – r}**× (**

**b)**

^{r}Now, on substituting **n = m, a =1 and b = 2x** we will get:

T_{r + 1} = ^{m}C_{r }× (1)^{m – r} × (2x)^{r}

_{r + 1 }= ^{m}C_{r} **× ****(2) ^{r} **

**×**

**(x)**

^{r}. . . . . . . . . . . . (1)Now, on comparing the coefficient of **x in equation (1) with x ^{2}**, we will get

**r = 2**

Therefore, from **equation (1):**

T_{ 2 + 1} = ^{m}C_{2} × (2)^{2} × (x)^{2}

Since, the coefficient of x^{2} = 140 **[GIVEN]**

Therefore, 140 = ^{m}C_{2} × (2)^{2}

Therefore, m^{2} – m – 42 = 0

Now, by **splitting of middle term** method, the roots of this quadratic equation are:

^{2} – (7 – 6)m – 42 = 0

^{2} – 7m + 6m – 42 = 0

i.e. (m + 6) (m – 7) = 0

**Therefore, m = 7 or m = -6**

Hence,** for the coefficient of x ^{2} in the expansion of (1+ 2x)^{k} to be 140,**

**the value of ‘m’ should be 7.**

** **

**Q.13: In the expansion of (2x+3y) ^{9}, Find the middle terms.**

**Sol.**

Here, **n = 9**

When **n** is **odd** then **the middle terms** in the expansion of **(a + b) ^{n}** are given by:

**middle terms =**

Therefore, the middle terms in the expansion of** (2x+3y) ^{9}**

**are:**

^{th} term and 6^{th} term

**Now, 5 ^{th} and 6^{th} terms in the expansion of **

**(2x+3y)**

^{9}are:**T _{5}** = T

_{4 + 1}=

^{9}C

_{4 }× (2x)

^{9 – 4}× (3y)

^{4}

_{5} = ^{9}C_{4 }× (2x)^{5} × (3y)^{4}

**Therefore, 5 ^{th} term = 414720 x^{5} y^{4}**

Now,** T _{6}** = T

_{5 + 1}=

^{9}C

_{5 }× (2x)

^{9 – 5}× (3y)

^{5}

_{6} = ^{9}C_{5 }× (2x)^{4} × (3y)^{5}

**Therefore, 6 ^{th} term = 489888 x^{4} y^{5}**

**Hence, 5 ^{th} and 6^{th} terms are the middle terms in the expansion of (2x+3y)^{9}**

**And also, 5 ^{th} term = 414720 x^{5} y^{4} and 6^{th} term = 489888 x^{4} y^{5}**

**Q.14: Find the Coefficient of x ^{6 }in expansion of (x + 3)^{11}**

**Sol.**

**Since, ****The general term in the expansion of (a + b) ^{n }is given by: T_{r + 1} = ^{n}C_{r }**

**×**

**(a)**

^{n – r}**×**

**b**

^{r}Therefore, on substituting **n = 11, a = x and b = 3** in the above expression we will get:

**T _{r + 1} = **

^{11}**C**

_{r }**×**

**(x)**

^{11 – r}**×**

**3**

^{r}. . . . . . . (1)Now, on comparing the coefficient of x in **equation (1) with x ^{7}**, we will get:

i.e. (x)^{11 – r} = x^{6}

(or) 11 – r = 6

**Therefore, r = 5**

Now, on substituting r in **equation (1)** we will get:

T_{5 + 1} = ^{11}C_{5 }× (x)^{11 – 5} × 3^{5}

**Therefore, ****the Coefficient of x ^{6 }in the expansion of (x + 3)^{11} = 112266**

** **

**Q.15: Write the general term in the expansion of (x ^{2}y^{3} – x^{3} y^{2})^{7}**

**Sol.**

The **general term** in the expansion of **(a + b) ^{n }**is given by

**: T**

_{r + 1}=^{n}C_{r }**×**

**(a)**

^{n – r}**×**

**b**

^{r}Now, on substituting **n = 7, a = ****x ^{2}y^{3}**

**and b =**

**x**we will get:

^{3}y^{2}**T _{r+1} = ^{7}C_{r }**

**×**

**(**

**x**

^{2}y^{3}**)**

^{7 – r}**× (x**

^{3}**y**

^{2})^{r}_{r + 1} = ^{7}C_{r }× (x^{2})^{7 – r} × (y^{3})^{7 – r} × (x)^{3 r }× (y)^{2 r}

_{r + 1} = ^{7}C_{r }× (x)^{14 – 2 r} × (x)^{3 r}× (y)^{2 r}× (y)^{21 – 3r}

_{r + 1} = ^{7}C_{r }× (x)^{14 – 2 r + 3r }× (y)^{21 – 3r + 2 r}

i.e. T_{r + 1} = ^{7}C_{r }× (x)^{14 + r }× (y)^{21 – r}

**Therefore, the general term in the ****expansion of (x ^{2}y^{3} – x^{3} y^{2})^{7}**

**:**

**T _{r + 1} = ^{7}C_{r }**

**×**

**(x)**

^{14 + r}**× (**

**y)**

^{21 – r}