Binomial Theorem

We know that-

Hence

Thus,

Putting a = 1 & b = (-2x), we get,

We know that-

Hence

Thus,

Putting a = (2/x) & b = (-x/2), we get-

We know that-

Hence

Thus,

Putting a = 2x & b = -3, we get-

We know that-

Hence

Thus,

Putting a = (x/3) & b = (1/x), we get-

We know that-

Hence

Thus,

Putting a = x & b = 1/x, we get-

(96)³ = (100-4)³

We know that-

Hence

Thus,

Putting a = 100 & b = -4, we get-

Thus,

(102)⁵ = (100+2)⁵

We know that-

Hence

Thus,

Putting a = 100 & b = 2, we get-

Hence, (102)⁵ = 11040808032

(101)⁴ = (100+1)⁴

We know that-

Hence

Thus,

Putting a = 100 & b = 1, we get-

Thus,

(99)⁵ = (100-1)⁵

We know that-

Hence

Thus,

Putting a = 100 & b = -1, we get-

Hence, (99)⁵ = 9509900499

(1.1)¹⁰⁰⁰⁰ = (1+0.1)¹⁰⁰⁰⁰

We know that-

Hence

Putting a = 1 & b = 0.1, we get-

Hence, (1.1)¹⁰⁰⁰⁰ is larger than 1000.

We know that-

Hence

Thus,

Replacing b with -b

Now,

Putting in , we get-

We know that-

Hence

Thus,

Replacing b with -b

Now,

Putting a = x & b = 1 in , we get-

putting x = √2

Hence,

In order to show that 9ⁿ⁺¹ – 8n – 9 is divisible by 64,

we have to prove that

9 ⁿ⁺¹ – 8n – 9 = 64 k, where k is some natural number

Now,

9ⁿ⁺¹ = (1+8)ⁿ⁺¹

We know that-

putting a =1, b = 8, and n = n+1

Hence,

Taking out (8)² from right side, we get-

where is a natural number

Thus, 9ⁿ⁺¹ – 8n – 9 is divisible by 64.

Hence Proved.

By Binomial Theorem

Putting b = 3 & a = 1 in the above equation, we get-

Hence Proved.

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

Here x⁵ is the T_r+1 term so

a= x, b = 3 and n =8

T_r+1 = ⁸C_r x^8-r 3^r……………1

For finding out x⁵

We equate x⁵= x^8-r

⇒ r= 3

Putting value of r in 1 we get

T₃₊₁ = ⁸C₃ x^8-3 3³

= 1512 x⁵

Hence the coefficient of x⁵= 1512

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

Here a =a , b = -2b & n =12

Putting values

T_r+1 = ¹²C_r a^12-r (-2b)^r……….1

To find a⁵

We equate a^12-r =a⁵

r=7

Putting r = 7 in 1

T₈ = ¹²C₇ a⁵ (-2b)⁷

= -101376 a⁵ b⁷

Hence the coefficient of a⁵b⁷= -101376

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r . ……..1

Here a = x2 , n =6 and b = -y

Putting values in 1

T_r+1 = ⁶C_r x ^2(6-r) (-1)^r y^r

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

Here n = 12, a= x² and b = -yx

Putting the values

T_n+1 =¹²C_r × x^2(12-r) (-1)^r y^r x^r

=

=

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

Here a= x , n =12 , r= 3 and b = -2y

T₄ = ¹²C₃ x⁹ (-2y)³

=

=

= -1760 x⁹ y³

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

Here a=9x , , n =18 and r = 12

Putting values

= (9 = (3²)⁶ = 3¹²)

= 18564

Here n = 7 so there would be two middle terms given by

Now

For T₄ , r= 3

The term will be

T_r+1 = ⁿC_r a^n-r b^r

For T₅ term , r = 4

The term T_r+1 in the binomial expansion is given by

T_r+1 = ⁿC_r a^n-r b^r

Here n is even so the middle term will be given by

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

Putting the values

=

= 61236 x⁵y⁵

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

Here n= m+n , a = 1 and b= a

Putting the values in the general form

T_r+1 = ^m+nC_r 1^m+n-r a^r

= ^m+nC_r a^r………….1

Now we have that the general term for the expression is,

T_r+1 = ^m+nC_r a^r
Now, For coefficient of a^m

T_m+1 = ^m+nCm a^m
Hence, for coefficient of a^m, value of r = m

So, the coefficient is ^m+nC_m

^m+nC_m =

And also,

^m+nC_{n =}

The coefficient of a^m and aⁿ are same i.e.;

Hence proved.

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

Here the binomial is (1+x)ⁿ with a = 1 , b = x and n = n

The (r+1)^th term is given by

T_(r+1) = ⁿC_r 1^n-r x^r

T_(r+1) = ⁿC_r x^r

The coefficient of (r+1)^th term is ⁿC_r

The r^th term is given by (r-1)^th term

T_(r+1-1) = ⁿC_r-1 x^r-1

T_r = ⁿC_r-1 x^r-1

∴ the coefficient of r^th term is ⁿC_r-1

For (r-1)^th term we will take (r-2)^th term

T_r-2+1 = ⁿC_r-2 x^r-2

T_r-1 = ⁿC_r-2 x^r-2

∴ the coefficient of (r-1)^th term is ⁿC_r-2

Given that the coefficient of (r-1)^th, r^th and r+1^th term are in ratio

1:3:5

⇒ 3r - 3 = n – r + 2

⇒ n - 4r + 5 =0…………1

Also

⇒

⇒

⇒

⇒

⇒

⇒

⇒ 5r = 3n - 3r + 3

⇒ 8r – 3n - 3 =0………….2

We have 1 and 2 as

n - 4r 5 =0…………1

8r – 3n - 3 =0…………….2

Multiplying equation 1 by number 2

2n -8r +10 =0……………….3

Adding equation 2 and 3

2n -8r +10 =0

+ -3n – 8r - 3 =0

⇒ -n = -7

n =7 and r = 3

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

The general term for binomial (1+x)²ⁿ is

T_r+1 = ²ⁿC_r x^r …………………..1

To find the coefficient of xⁿ

r=n

T_n+1 = ²ⁿC_n xⁿ

The coefficient of xⁿ = ²ⁿC_n

The general term for binomial (1+x)^2n-1 is

T_r+1 = ^2n-1C_r x^r

To find the coefficient of xⁿ

Putting n =r

T_r+1 = ^2n-1C_r xⁿ

The coefficient of xⁿ = ^2n-1C_n

We have to prove

Coefficient of xⁿ in (1+x)²ⁿ = 2 coefficient of xⁿ in (1+x)^2n-1

Hence L.H.S = R.H.S.

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

Here a = 1, b = x and n = m

Putting the value

T_r+1 = ^mC_r 1^m-r x^r

= ^mC_r x^r

We need coefficient of x²

∴ putting r = 2

T₂₊₁ = ^mC₂ x²

The coefficient of x² = ^mC₂

Given that coefficient of x² = ^mC₂ = 6

⇒

⇒

⇒ m(m-1) = 12

⇒ m²- m - 12 =0

⇒ m²- 4m +3m - 12 =0

⇒ m(m-4) + 3(m-4) = 0

⇒ (m+3)(m - 4)= 0

⇒ m = - 3, 4

we need positive value of m so m = 4

We know that-

So first 3 terms are

Also, it is given that their value are 729, 7290 and 30375

Therefore,

…(1)

…(2)

…(3)

Dividing (1) by (2)

…(4)

Dividing (2) by (3)

…(5)

Dividing (4) by (5)

⇒ 12(n-1) = 10n

⇒ 12n-12 = 10n

⇒ 2n = 12

⇒ n = 12/2

∴ n = 6

Putting n = 6 in (1)

aⁿ = 729

⇒ a⁶ = 729

⇒ a⁶ = (3)⁶

∴ a = 3

Putting a = 3, n = 6 in (5)

⇒ 6b = 30

⇒ b = 30/6

∴ b = 5

Thus, a = 3, b = 5 and n = 6.

We know that-

General term of expansion (a+b)ⁿ is

For (3+ax)⁹

Putting a = 3, b = ax & n = 9

General term of (3+ax)⁹ is

Since we need to find the coefficients of x² and x³, therefore

For r = 2

Thus, the coefficient of x² =

For r = 3

Thus, the coefficient of x³ =

Given that-

Coefficient of x² = Coefficient of x³

∴ a = 9/7

Hence, a = 9/7

We know that-

Hence

= a⁶ + 6a⁵b + 15a⁴b² + 20a³b³ + 15a²b⁴ + 6ab⁵ + b⁶

Thus, (a + b)⁶ = a⁶ + 6a⁵b + 15a⁴b² + 20a³b³ + 15a²b⁴ + 6ab⁵ + b⁶

Putting a = 1 & b = 2x, we get-

(1+2x)⁶ = (1)⁶ + 6(1)⁵(2x) + 15(1)⁴(2x)² + 20(1)³(2x)³

+ 15(1)²(2x)⁴ + 6(1)(2x)⁵ + (2x)⁶

= 1 + 12x + 60x² + 160x³ + 240x⁴ + 192x⁵ + 64x⁶

Similarly,

= a⁷ + 7a⁶b + 21a⁵b² + 35a⁴b³ + 35a³b⁴ + 21a²b⁵ + 7ab⁶ +b⁷

Thus, (a + b)⁷ = a⁷ + 7a⁶b + 21a⁵b² + 35a⁴b³ + 35a³b⁴ + 21a²b⁵ + 7ab⁶ +b⁷

Putting a = 1 & b = -x, we get-

(1-x)⁷ = (1)⁷ + 7(1)⁶(-x) + 21(1)⁵(-x)² + 35(1)⁴(-x)³

+ 35(1)³(-x)⁴ + 21(1)²(-x)⁵ + 7(1)(-x)⁶ + (-x)7

= 1 - 7x + 21x² - 35x³ + 35x⁴ - 21x⁵ + 7x⁶ - x⁷

Now,

(1+2x)⁶(1-x)⁷

=(1 + 12x + 60x² + 160x³ + 240x⁴ + 192x⁵ + 64x⁶)

× (1 - 7x + 21x² - 35x³ + 35x⁴ - 21x⁵ + 7x⁶ - x⁷)

Coefficient of x⁵ = [(1)×(-21) + (12)×(35) + (60)×(-35)

+ (160)×(21) + (240)×(-7) + (192)×(1)]

= [-21 + 420 - 2100 + 3360 - 1680 + 192]

= 171

Thus, the coefficient of x⁵ in the expression (1+2x)⁶(1-x)7 is 171.

We can write aⁿ as

aⁿ = (a-b+b)ⁿ

We know that-

putting a = b & b = a-b, we get-

where

Hence (a-b) is a factor of (aⁿ-bⁿ).

We know that-

Hence

= a⁶ + 6a⁵b + 15a⁴b² + 20a³b³ + 15a²b⁴ + 6ab⁵ + b⁶

Thus, (a + b)⁶ = a⁶ + 6a⁵b + 15a⁴b² + 20a³b³ + 15a²b⁴ + 6ab⁵ + b⁶

Replacing b with -b

(a +(-b))⁶ = a⁶ + 6a⁵(-b) + 15a⁴(-b)² + 20a³(-b)³ + 15a²(-b)⁴ + 6a(-b)⁵ + (-b)⁶

(a - b)⁶ = a⁶ - 6a⁵b + 15a⁴b² - 20a³b³ + 15a²b⁴ - 6ab⁵ + b⁶

Now,

(a + b)⁶ - (a - b)⁶

= (a⁶ + 6a⁵b + 15a⁴b² + 20a³b³ + 15a²b⁴ + 6ab⁵ + b⁶)

- (a⁶ - 6a⁵b + 15a⁴b² - 20a³b³ + 15a²b⁴ - 6ab⁵ + b⁶)

= 2(6a⁵b + 20a³b³ + 6ab⁵)

Putting a = √3 & b = √2, we get-

(√3 + √2)⁶ - (√3 - √2)⁶

= 2[6(√3)⁵(√2) + 20(√3)³(√2)³ + 6(√3)(√2)⁵]

= 2[6(9√3)(√2) + 20(3√3)(2√2) + 6(√3)(4√2)]

= 2[54√6 + 120√6 + 24√6]

= 2[198√6]

= 396√6

We know that-

Hence

= a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴

Thus, (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴

Replacing b with -b

(a + (-b))⁴ = a⁴ + 4a³(-b) + 6a²(-b)² + 4a(-b)³ + (-b)⁴

(a-b)⁴ = a⁴ - 4a³b + 6a²b² - 4ab³ + b⁴

Now,

(a + b)⁴ + (a - b)⁴

= (a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴) + (a⁴ - 4a³b + 6a²b² - 4ab³ + b⁴)

= 2(a⁴ + 6a²b² + b⁴)

Putting a = a² & b = √(a²-1)

(a² + √(a²-1))⁴ + (a² - √(a²-1))⁴

= 2[(a²)⁴ + 6(a²)²(√(a²-1))² + (√(a²-1))⁴]

= 2[ a⁸ + 6(a⁴)(a²-1) + (a²-1)²]

= 2[ a⁸ + 6 a⁶ - 6a⁴ + a⁴ - 2 a² + 1]

= 2[ a⁸ + 6 a⁶ - 5a⁴ - 2 a² + 1]

(0.99)⁵ = (1 - 0.01)⁵

We know that-

Hence

= a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵

Thus, (a + b)⁵ = a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵

Putting a = 1 & b = -0.01, we get-

(1 + (-0.01))⁵ = (1)⁵ + 5(1)⁴(-0.01) + 10(1)³(-0.01)² + 10(1)²(-0.01)³ + 5(1)(-0.01)⁴ + (-0.01)⁵

Using first three terms,

(0.99)⁵ = (1)⁵ + 5(1)⁴(-0.01) + 10(1)³(-0.01)²

= 1 - 0.05 + 0.001

= 1.001 - 0.050

= 0.951

We know that

General term of expansion (a + b)ⁿ

We need to calculate fifth term from beginning of expansion

∴ putting r = 4, a = , and b = , we get-

Now,

In the expression of (a + b)ⁿ

r^th term from the end = (n-r+2)^th term from the begining

Hence, 5th term from the end

= (n-5+2)^th term from the beginning

= (n-3)^th term from the beginning

Now, We need to calculate (n-3)^th term from beginning of expansion

putting r = (n-3)-1 = n-4, a = , and b = , we get-

Given that-

Comparing powers of 6

⇒ 2(n-8) = 4

⇒ 2n-16 = 4

⇒ 2n = 20

∴ n = 20/2 = 10

Thus, the value of n is 10.

We know that-

Hence

= a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴

Thus, (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴ …(1)

Putting , we get-

…(2)

Now Solving separately

From (1)

(a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴

putting a = 1 & b = (x/2), we get-

we know that-

(a + b)³ = a³ + 3a²b + 3ab² + b³

putting a = 1 & b = (x/2), we get-

Substituting the value of in (2), we get-

Thus,

We know that-

(a + b)³ = a³ + 3a²b + 3ab² + b³

putting a = 3x² & b = -a(2x-3a), we get-

[3x² + (-a(2x-3a))]³

= (3x²)³+3(3x²)²(-a(2x-3a)) + 3(3x²)(-a(2x-3a))² + (-a(2x-3a))³

= 27x⁶ - 27ax⁴(2x-3a) + 9a²x²(2x-3a)² - a³(2x-3a)³

= 27x⁶ - 54ax⁵ + 81a²x⁴ + 9a²x²(4x²-12ax+9a²)

- a³[(2x)³ - (3a)³ - 3(2x)²(3a) + 3(2x)(3a)²]

= 27x⁶ - 54ax⁵ + 81a²x⁴ + 36a²x⁴ - 108a³x³ + 81a⁴x²

-8a³x³ + 27a⁶ + 36a⁴x² - 54a⁵x

= 27x⁶ - 54ax⁵+ 117a²x⁴ - 116a³x³ + 117a⁴x² - 54a⁵x + 27a⁶

Thus, (3x² – 2ax + 3a²)³

= 27x⁶ - 54ax⁵+ 117a²x⁴ - 116a³x³ + 117a⁴x² - 54a⁵x + 27a⁶