Class 11 Maths Ncert Solutions Chapter 8 Ex 8.2 Binomial Theorem PDF

Class 11 Maths Ncert Solutions Ex 8.2

Class 11 Maths Ncert Solutions Chapter 8 Ex 8.2

Formulas:

 

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

      Tr + 1 = nCr × (a)n – r × br

 

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 x6 in the expansion of (x + 2)9

 

Sol.

Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

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

Tr + 1 = 9Cr × (x)9 – r × 2r . . . . . . . (1)

Now, on comparing the coefficient of x in equation (1) with x6, we will get:

i.e.         (x)9 – r = x6                                    

(or)         9 – r = 6

Therefore, r = 3

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

T4 = 9C3 × (x)9 – 3 × 23

T4=9!3!6!×x6×8 T4=9×8×7×6!3×2×1×6!×8x6=672x6

Therefore, the Coefficient of x6 in the expansion of (x + 2)9 = 672

 

 

Q.2: Find the Coefficient of a6 b8 in the expansion of (2a – 3b)14

 

Sol.

Since, The general term in the expansion of (x + y)n is given by: Tr + 1 = nCr × (x)n – r × yr

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

Tr + 1 = 14Cr × (2a)14 – r × (-3b)r

i.e.    Tr + 1 = [14Cr × (2)14 – r (-3)r ] (a)14 – r × (b)r  . . . . . . . . . .  (1)

Now, on comparing the coefficients of a and b in equation (1) with a6 b8, we will get:

i.e.         a6 b8 = a14 – r br

Therefore,        r = 8

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

T9 = [ 14C8 × (2)14 – 8 (-3)8 ] (a)14 – 8 × (b)8

i.e.     T9 = [ 14C8 × (2)6 (-3)8 ] a6 b8

T9=14!8!6!×64×729×a6b8 T9=14×13×12×11×10×9×8!6×5×4×3×2×1×8!×64×729×a6b8 T9=3003×64×729×a6b8=140107968a6b8

Therefore, the Coefficient of a6 b8 in the expansion of (2a – 3b)14 = 140107968

 

 

Q.3: Write the general term in the expansion of (x3 – y2)5

 

Sol.

The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Now, on substituting n = 5, a = x3 and b = y2 we will get:

Tr+1 = 5Cr × (x3)5 – r × (y2)r

Therefore, the general term in the expansion of (x3 – y2)5:

Tr+1 = 5Cr × (x)15 – 3 r × (y)2 r

 

 

Q.4: Write the general term in the expansion of (x4 – xy2)9

 

Sol.

The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Now, on substituting n = 9, a = x4 and b = xy2 we will get:

Tr+1 = 9Cr × (x4)9 – r × (xy2)r

Tr+1 = 9Cr × (x)36 – 4 r × (x)2 r × (y)2 r

Tr+1 = 9Cr × (x)36 – 4 r + 2 r × (y)2 r

Tr+1 = 9Cr × (x)36 – 2 r × (y)2 r

Therefore, the general term in the expansion of (x4 – xy2)9:

Tr+1 = 9Cr × (x)36 – 2 r × (y)2r

 

 

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

 

Sol.

Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Therefore, from the above equation the value of r should be 3; for finding out the values of 4th Term (T4)

Therefore, T3 + 1 = nC3 × (a)n – 3 × b3

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

   T3 + 1 = 10C3 × (2x)10 – 3 × (3y)3

   T4 = 10C3 × (2x)7 × (3y)3

T4=10!3!7!×128x7×27y3 T4=10×9×8×7!3×2×1×7!×3456x7y3 T4=120×3456x7y3=414720x7y3

Therefore, 4th term in the expansion of (2x + 3y)10 = 414720 x7y3

 

 

Q.6 Find the 11th term in the expansion of (8x12x)15

                                                                                                     

Sol.

Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Therefore, from the above equation the value of r should be 10 for finding out the values of 11th Term (T11)

Therefore, T10 + 1 = nC10 × (a)n – 10 × b10

Now, on substituting n =15, a = 8x and b = (12x) we will get:

   T10 + 1 = 15C10 × (8x)15 – 10 × (12x)10

   T11 = 15C10 × (8x)5 × (12x)10

T11=15!10!5!×8×8×8×8×8×x5×1210×(1x)10 T11=15×14×13×12×11×10!5×4×3×2×1×10!×32×x5×1x5 T11=3003×32=96096

Therefore, 11th term in the expansion of (8x12x)15= 96096

 

 

Q.7: In the expansion of (5x410)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 = (n+12)thtermand(n+12+1)thterm

Therefore, the middle terms in the expansion of (5x410)9are:

(9+12)thtermand(9+12+1)thterm

5th term and 6th term

Now, 5th and 6th terms in the expansion of (5x410)9 are:

T5 = T4 + 1 = 9C4 × (5)9 – 4 × (x410)4

   T5 = 9C4 × (5)5 × (x410)4

T5=9!4!5!×(5)5×(x410)4 T5=9×8×7×6×5!4×3×2×1×5!×5×5×5×5×5×1104×x16 T5=126×516x16=3158x16

Therefore, 5th term =3158x16

Now, T6 = T5 + 1 = 9C5 × (5)9 – 5 × (x410)5

   T6 = 9C5 × (5)4 × (x410)4

T6=9!5!4!×(5)4×(x410)5 T6=9×8×7×6×5!5!×(4×3×2×1)×5×5×5×5×1105×x20 T6=126×1320x20=63160x20

Therefore, 6th term =63160x20

Therefore, 5th and 6th terms are the middle terms in the expansion of (5x410)9

And also, 5th term =3158x16 and 6th 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:

(n2+1)thterm

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

(102+1)thterm = 6th term

Now, 6th term in the expansion of (x2+8y)10 is:

T6 = T5 + 1 = 10C5 × (x2)105 × (8y)5

   T6 = 10C5 × (x2)5 × (8y)5

T6=10!5!5!×(x2)5×(8y)5 T6=10×9×8×7×6×5!5×4×3×2×1×5!×x532×8×8×8×8×8×y5 T6=252×1024x5y5=258048x5y5

Therefore, 6th term = 258048 x5 y5

Hence, the middle term in the expansion of (x2+8y)10 = 258048 x5 y5

 

 

Q.9: Prove that the coefficients of pand pb 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: Tr + 1 = nCr × (x)n – r × (y)r

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

Tr + 1 = (a + b)Cr × (1)(a + b) – r × (p)r

Tr + 1 = (a + b)Cr × (p)r . . . . . . . . . . . . (1)

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

Therefore, from equation (1):

T a + 1 = (a + b)Ca × (p)a

Ta+1=(a+b)!a!(a+ba)!×pa Ta+1=(a+b)!a!b!×pa

Therefore, the coefficient of (p)a =  (a+b)!a!b! . . . . . . . . . . (2)

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

Therefore, from equation (1):

T b + 1 = (a + b)Cb × (p)b

Tb+1=(a+b)!b!(a+bb)!×pb Tb+1=(a+b)!b!a!×pb

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 pand pb are equal in the expansion of (1 + p)a + b.

Hence, Proved

 

 

Q.10: The coefficients of the (k + 1)th, kth 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: Tr + 1 = nCr × (a)n – r × br

Now on substituting a = x and b = 1 in the above equation:

Tr + 1 = nCr × (x)n – r × (1)r

Now, (k + 1)th term in the expansion of (x + 1)n :

T(k) + 1 = nCk × (x)n – k × 1(k )

i.e.  T(k + 1) = nCk × (x)n – k

Tk+1=n!k!(nk)!×(x)nk

Therefore the coefficient of T(k + 1)th term = n!k!(nk)!

Now, (k)th term in the expansion of (x + 1)n :

T(k – 1) + 1 = nCk – 1 × (x)n – (k – 1) × 1(k – 1)

i.e.  Tk = nCk – 1 × (x)n – k + 1

Tk=n!(k1)!(nk+1)!×(x)nk+1

Therefore the coefficient of  T(k)th term = n!(k1)!(nk+1)!

And, (k – 1)th term in the expansion of (x + 1)n :

T(k – 2 )+ 1 = nCk – 2 × (x)n – (k – 2) × 1(k – 2)

i.e.  Tk – 1 = nCk – 2 × (x)n – k + 2

Tk1=n!(k2)!(nk+2)!×(x)nk+2

Therefore the coefficient of  T(k – 1)th term = n!(k2)!(nk+2)!

Now, according to the given conditions: Coefficients of T(k + 1), Tk and Tk – 1 are in the ratio of 5 : 3 : 1

Therefore, Tk1Tk=13

[n!(k2)!(nk+2)!]÷[n!(k1)!(nk+1)!]=13 (k1)!(nk+1)!(k2)!(nk+2)!=13 (k1)(k2)!(nk+1)!(k2)!(nk+2)(nk+1)!=13 k1nk+2=13 3k3=nk+2

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

And,        Tk+1Tk=53

[n!k!(nk)!]÷[n!(k1)!(nk+1)!]=53 (k1)!(nk+1)!k!(nk)!=53 (k1)![(nk+1)(nk)!][k(k1)!](nk)!=53 nk+1k=53 3n3k+3=5k

(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

3n – 8k + 3 – 3n + 12k – 15 = 0

4k = 12

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 ap in the expansion of (1 + 3a)2p – 1 is half the coefficient of ap in the expansion of (1 + 3a)2p

 

Sol.

Since, The general term in the expansion of (x + y)n is given by: Tr + 1 = nCr × (x)n – r × (y)r

For (1 + 3a)2p:

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

Tr + 1 = (2p)Cr × (1)(3p) – r × (3a)r

Tr + 1 = (2p)Cr × (3a)r . . . . . . . . . . . . (1)

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

Therefore, from equation (1):

T p + 1 = (2p)Cp × (3a)p

Tp+1=(2p)!p!(2pp)!×(3a)p

Tp+1=(2p)!p!p!×3p×(a)p

Therefore, the coefficient of ap =(2p)!p!p!×3p . . . . . . . . . . . . . . . . (2)

Now, for (1 + 3a)2p – 1 :

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

Tr + 1 = (2p – 1)Cr × (1)(2p – 1) – r × (3a)r

Tr + 1 = (2p – 1)Cr × (3a)r . . . . . . . . . . . . (1)

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

Therefore, from equation (1):

T p + 1 = (2p – 1)Cp × (3a)p

Tp+1=(2p1)!p!(2p1p)!×(3a)p Tp+1=(2p1)!p!(p1)!×(3)p×(a)p Tp+1=(2p)!2pp!×p!p×(3)p.(a)p=(2p)!2(p!)2×3p.ap

Therefore, the coefficient of ap = =(2p)!2(p!)2×3p . . . . . . . . . . (3)

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

(2p)!p!p!×3p=(2p)!p!p!×3p

  (2p)Cp × 312 (2p – 1)Cp × 3p

Therefore, the coefficient of ap in the expansion of (1 + 3a)2p – 1 is half the coefficient of ap in the expansion of (1 + 3a)2p

 

 

Q.12: For what values of ‘m’ the coefficient of x2 in the expansion of (1+ 2x)k is 140.

 

Sol.

Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × (b)r

Now, on substituting n = m, a =1 and b = 2x we will get:

Tr + 1 = mCr × (1)m – r × (2x)r

Tr + 1 = mCr × (2)r × (x)r . . . . . . . . . . . . (1)

Now, on comparing the coefficient of x in equation (1) with x2, we will get r = 2

Therefore, from equation (1):

T 2 + 1 = mC2 × (2)2 × (x)2

Since, the coefficient of x2 = 140   [GIVEN]

Therefore,    140 = mC2 × (2)2

84=[m!2!(m2)!]×4 21=m×(m1)(m2)![2×1]×(m2)! 42=m2m

Therefore, m2 – m – 42 = 0

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

m2 – (7 – 6)m – 42 = 0

m2 – 7m + 6m – 42 = 0

m(m – 7) +6(m – 7) = 0

i.e.    (m + 6) (m – 7) = 0

Therefore, m = 7 or m = -6

Hence, for the coefficient of x2 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 = (n+12)thtermand(n+12+1)thterm

Therefore, the middle terms in the expansion of (2x+3y)9 are:

(9+12)thtermand(9+12+1)thterm

5th term and 6th term

Now, 5th and 6th terms in the expansion of (2x+3y)9 are:

T5 = T4 + 1 = 9C4 × (2x)9 – 4 × (3y)4

   T5 = 9C4 × (2x)5 × (3y)4

T5=9!4!5!×(2x)5×(3y)4 T5=9×8×7×6×5!4×3×2×1×5!×32x5×81y4

T5=126×2592x5y4=414720x5y4

Therefore, 5th term = 414720 x5 y4

Now, T6 = T5 + 1 = 9C5 × (2x)9 – 5 × (3y)5

T6 = 9C5 × (2x)4 × (3y)5

T6=9!5!×4!×(2x)4×(3y)5 T6=9×8×7×6×5!5!×(4×3×2×1)×16x4×243y5 T6=126×3888x4y5=489888x4y5

Therefore, 6th term = 489888 x4 y5

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

And also, 5th term = 414720 x5 y4 and 6th term = 489888 x4 y5

 

 

Q.14: Find the Coefficient of x6 in expansion of (x + 3)11

 

Sol.

Since, The general term in the expansion of (a + b)is given by:  Tr + 1 = nCr × (a)n – r × br

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

Tr + 1 = 11Cr × (x)11 – r × 3r . . . . . . . (1)

Now, on comparing the coefficient of x in equation (1) with x7, we will get:

i.e.         (x)11 – r = x6

(or)         11 – r = 6

Therefore, r = 5

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

T5 + 1 = 11C5 × (x)11 – 5 × 35

T6=11!5!×6!×x6×35 T6=11×10×9×8×7×6!5×4×3×2×1×6!×243x6

T6=462×243x6=112266x6

Therefore, the Coefficient of x6 in the expansion of (x + 3)11 = 112266

 

 

Q.15: Write the general term in the expansion of (x2y3 – x3 y2)7

 

Sol.

The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Now, on substituting n = 7, a = x2y3 and b = x3 y2 we will get:

Tr+1 = 7Cr × (x2y3)7 – r × (x3 y2)r

Tr + 1 = 7Cr × (x2)7 – r × (y3)7 – r × (x)3 r × (y)2 r

Tr + 1 = 7Cr × (x)14 – 2 r × (x)3 r× (y)2 r× (y)21 – 3r

Tr + 1 = 7Cr × (x)14 – 2 r + 3r  ×  (y)21 – 3r + 2 r

i.e.  Tr + 1 = 7Cr × (x)14 + r  ×  (y)21 – r

Therefore, the general term in the expansion of (x2y3 – x3 y2)7:

Tr + 1 = 7Cr × (x)14 + r × (y)21 – r

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