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(i) Let us check commutativity of *

Let a, b ∈ Q, then

a * b = ab²

b * a = ba²

So,

a * b ≠ b * a

Thus, * is not commutative on Q

Let us check the associativity of *

Let a, b, c ∈ Q, then

a * (b * c) = a * (bc²)

= a(bc²)²

= ab²c⁴

(a * b) * c = (ab²) * c

= ab²c²

So, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(ii) We have to check commutativity of *

Let a, b ∈ Q, then

a * b = a + ab

b * a = b + ba

= b + ab

So, a * b ≠ b * a

Thus, * is not commutative on Q.

Let us prove associativity on Q.

Let a, b, c ∈ Q, then

a * (b * c) = a * (b + b c)

= a + a (b + b c)

= a + ab + a b c

(a * b) * c = (a + a b) * c

= (a + a b) + (a + a b) c

= a + a b + a c + a b c

So, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(iii) Let us check the commutativity of *

Let a, b ∈ R, then

a * b = a + b – 7

= b + a – 7

= b * a

Therefore,

a * b = b * a, for all a, b ∈ R

Thus, * is commutative on R

Let us prove associativity of * on R.

Let a, b, c ∈ R, then

a * (b * c) = a * (b + c – 7)

= a + b + c - 7 - 7

= a + b + c – 14

(a * b) * c = (a + b – 7) * c

= a + b – 7 + c – 7

= a + b + c – 14

So,

a * (b * c ) = (a * b) * c, for all a, b, c ∈ R

Thus, * is associative on R.

(iv) Let us check commutativity of *

Let a, b ∈ Q, then

a * b = (a – b)²

= (b – a)²

= b * a

Therefore,

a * b = b * a, for all a, b ∈ Q

Thus, * is commutative on Q

Let us prove the associativity of * on Q

Let a, b, c ∈ Q, then

a * (b * c) = a * (b – c)²

= a * (b² + c² – 2 b c)

= (a – b² – c² + 2bc)²

(a * b) * c = (a – b)² * c

= (a² + b² – 2ab) * c

= (a² + b² – 2ab – c)²

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(v) Let us check the commutativity of *

Let a, b ∈ Q, then

a * b = ab + 1

= ba + 1

= b * a

Therefore

a * b = b * a, for all a, b ∈ Q

Thus, * is commutative on Q

Let us prove the associativity of * on Q

Let a, b, c ∈ Q, then

a * (b * c) = a * (bc + 1)

= a (b c + 1) + 1

= a b c + a + 1

(a * b) * c = (ab + 1) * c

= (ab + 1) c + 1

= a b c + c + 1

So, a * (b * c) ≠ (a * b) * c

Hence, * is not associative on Q.

(i) Given ‘*’ on N defined by a * b = a^b for all a, b ∈ N.

Let a, b ∈ N. Then,

a^b ∈ N [∵ a^b ≠ 0 and a, b is positive integer]

⇒ a * b ∈ N

So,

a * b ∈ N, ∀ a, b ∈ N

Thus, * is a binary operation on N.

(ii) Given ‘O’ on Z defined by a O b = a^b for all a, b ∈ Z.

Both a = 3 and b = -1 belong to Z.

⇒ a * b = 3^-1

= 1/3 ∉ Z

Thus, * is not a binary operation on Z.

(iii) Given ‘*’ on N defined by a * b = a + b – 2 for all a, b ∈ N

If a = 1 and b = 1,

a * b = a + b – 2

= 1 + 1 – 2

= 0 ∉ N

Thus, there exist a = 1 and b = 1 such that a * b ∉ N

Therefore, * is not a binary operation on N.

(i) Let X = (a, b) and Y = (c, d) ∈ A, ∀ a, c ∈ R₀ and b, d ∈ R

Then, X O Y = (ac, bc + d)

And Y O X = (ca, da + b)

S0,

X O Y = Y O X, ∀ X, Y ∈ A

Thus, O commutative on A.

Let us check the associativity of O

Let X = (a, b), Y = (c, d) and Z = (e, f), ∀ a, c, e ∈ R₀ and b, d, f ∈ R

X O (Y O Z) = (a, b) O (ce, de + f)

= (ace, bce + de + f)

(X O Y) O Z = (ac, bc + d) O (e, f)

= (ace, (bc + d) e + f)

= (ace, bce + de + f)

So, X O (Y O Z) = (X O Y) O Z, ∀ X, Y, Z ∈ A

(ii) Let E = (x, y) be the identity element in A with respect to O, ∀ x ∈ R₀ and y ∈ R

Such that,

XOE = X = EOX, ∀ X ∈ A

XOE = X and EOX = X

(ax, bx +y) = (a, b) and (xa, ya + b) = (a, b)

Consider (ax, bx + y) = (a, b)

ax = a

x = 1

And bx + y = b

y = 0 [since x = 1]

Consider (xa, ya + b) = (a, b)

xa = a

x = 1

And ya + b = b

y = 0 [since x = 1]

So, (1, 0) is the identity element in A with respect to O.

(iii) Let F = (m, n) be the inverse in A ∀ m ∈ R₀ and n ∈ R

X O F = E and F O X = E

(am, bm + n) = (1, 0) and (ma, na + b) = (1, 0)

Consider (am, bm + n) = (1, 0)

am = 1

m = 1/a

And bm + n = 0

n = -b/a [since m = 1/a]

Consider (ma, na + b) = (1, 0)

ma = 1

m = 1/a

And na + b = 0

n = -b/a

So, the inverse of (a, b) ∈ A with respect to O is (1/a, -1/a)

(i) Given that ‘×₆‘ on S = {1, 2, 3, 4, 5} defined by a × ₆b = Remainder when ab is divided by 6.

Consider the table,

Here all elements of table are not in S.

⇒ For a = 2 and b = 3,

a × ₆b = 2 × ₆3 = remainder when 6 divided by 6 = 0 ≠ S

So, ×₆ is not a binary operation on S.

(ii) Given ‘+₆’ on S = {0, 1, 2, 3, 4, 5} defined by a + ₆b

= {(a + b, if a + b < 6), (a + b - 6, if a + b ≥ 6)

Consider the table

Here all elements of table are not in S.

⇒ For a = 2 and b = 3,

a × ₆b = 2 × ₆3 = remainder when 6 divided by 6 = 0 ≠ Thus, ×₆ is not a binary operation on S.

(iii) Given that ‘⊙’ on N defined by a ⊙ b = a^b + b^a for all a, b ∈ N

Let a, b ∈ N. Then,

a^b, b^a ∈ N

⇒ a^b + b^a ∈ N [∵Add in binary operation on N]

⇒ a ⊙ b ∈ N

So, ⊙ is a binary operation on N.

(iv) Given ‘*’ on Q defined by a * b = (a – 1)/(b + 1) for all a, b ∈ Q

If a = 2 and b = -1 in Q,

a * b = (a – 1)/(b + 1)

= (2 – 1)/(- 1 + 1)

= 1/0 [which is not defined]

For a = 2 and b = -1

a * b does not belongs to Q

Therefore, * is not a binary operation in Q.

The number of binary operations on a set with n elements is n^{n^2}

Here, S = {a, b, c}

The number of elements in S = 3

The number of binary operations on a set with 3 elements is 3^{3^2}

(i) First Let us check commutativity of *

Let a, b ∈ Z

Then a * b = a + b + ab

= b + a + ba

= b * a

So,

a * b = b * a, ∀ a, b ∈ Z

Let us prove associativity of *

Let a, b, c ∈ Z, Then,

a * (b * c) = a * (b + c + bc)

= a + (b + c + bc) + a (b + c + bc)

= a + b + c + bc + ab + ac + abc

(a * b) * c = (a + b + ab) * c

= a + b + a b + c + (a + b + a b) c

= a + b + a b + c + a c + b c + a b c

So,

a * (b * c) = (a * b) * c, ∀ a, b, c ∈ Z

Thus, * is associative on Z.

(ii) Let us check commutativity of *

Let a, b ∈ N

a * b = 2^ab

= 2^ba

= b * a

So, a * b = b * a, ∀ a, b ∈ N

Thus, * is commutative on N

Let us check associativity of *

Let a, b, c ∈ N

Then, a * (b * c) = a * (2^bc)

=2a ∗ 2bc

(a * b) * c = (2^ab) * c

=2ab ∗ 2c

So, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on N

(iii) Let us check commutativity of *

Let a, b ∈ Q, then

a * b = a – b

b * a = b – a

So, a * b ≠ b * a

Thus, * is not commutative on Q

Let us check associativity of *

Let a, b, c ∈ Q, then

a * (b * c) = a * (b – c)

= a – (b – c)

= a – b + c

(a * b) * c = (a – b) * c

= a – b – c

Therefore,

a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q

(iv) Let us check commutativity of ⊙

Let a, b ∈ Q, then

a ⊙ b = a² + b²

= b² + a²

= b ⊙ a

So, a ⊙ b = b ⊙ a, ∀ a, b ∈ Q

Thus, ⊙ on Q

Let us check associativity of ⊙

Let a, b, c ∈ Q, then

a ⊙ (b ⊙ c) = a ⊙ (b² + c²)

= a² + (b² + c²)²

= a² + b⁴ + c⁴ + 2b²c²

(a ⊙ b) ⊙ c = (a² + b²) ⊙ c

= (a² + b²)² + c²

= a⁴ + b⁴ + 2a²b² + c²

So,

(a ⊙ b) ⊙ c ≠ a ⊙ (b ⊙ c)

Thus, ⊙ is not associative on Q.

(v) Let us check commutativity of o

Let a, b ∈ Q, then

a o b = (ab/2)

= (ba/2)

= b o a

So, a o b = b o a, ∀ a, b ∈ Q

Thus, o is commutative on Q

Let us check associativity of o

Let a, b, c ∈ Q, then

a o (b o c) = a o (b c/2)

= [a (bc/2)]/2

= [a (bc/2)]/2

= (a b c)/4

(a o b) o c = (ab/2) o c

= [(ab/2)c]/2

= (a b c)/4

So, a o (b o c) = (a o b) o c, ∀ a, b, c ∈ Q

Hence, o is associative on Q.

(i) Given that On^Z+, defined * by a * b = a – b

If a = 1 and b = 2 in Z⁺, then

a * b = a – b

= 1 – 2

= -1 ∉ Z⁺ [because Z⁺ is the set of non-negative integers]

For a = 1 and b = 2,

a * b ∉ Z⁺

So, * is not a binary operation on Z⁺.

(ii) Given that Z⁺, define * by a*b = ab

Let a, b ∈ Z⁺

⇒ a, b ∈ Z⁺

⇒ a * b ∈ Z⁺

So, * is a binary operation on R.

(iii) Given that on R, define by a*b = ab²

Let a, b ∈ R

⇒ a, b² ∈ R

⇒ ab² ∈ R

⇒ a * b ∈ R

So, * is a binary operation on R.

(iv) Given that on Z⁺ define * by a * b = |a − b|

Let a, b ∈ Z⁺

⇒ |a – b| ∈ Z⁺

⇒ a * b ∈ Z⁺

So,

a * b ∈ Z⁺, ∀ a, b ∈ Z⁺

Thus, * is a binary operation on Z⁺.

(v) Given that on Z⁺ define * by a * b = a

Let a, b ∈ Z⁺

⇒ a ∈ Z⁺

⇒ a * b ∈ Z⁺

So, a * b ∈ Z⁺ ∀ a, b ∈ Z⁺

Thus, * is a binary operation on Z⁺.

(vi) Given that On R, define * by a * b = a + 4b²

Let a, b ∈ R

⇒ a, 4b² ∈ R

⇒ a + 4b² ∈ R

⇒ a * b ∈ R

So, a *b ∈ R, ∀ a, b ∈ R

Hence, * is a binary operation on R.

Let a, b ∈ A

Then, a * b = b

b * a = a

So, a * b ≠ b * a

Thus, * is not commutative on A

Let us check associativity:

Let a, b, c ∈ A

a * (b * c) = a * c

= c

So

a * (b * c) = (a * b) * c, ∀ a, b, c ∈ A

Hence, * is associative on A

Consider the composition table :

In the given composition table, all the elements are not in the set {1, 2, 3, 4, 5}.

If we consider a = 2 and b = 3, a * b = LCM of a and b = 6 ∉ {1, 2, 3, 4, 5}.

Thus, * is not a binary operation on {1, 2, 3, 4, 5}.

Given that a * b = 2a + b – 3

3 * 4 = 2(3) + 4 – 3

= 6 + 4 – 3

= 7

Let a, b ∈ Q – {-1}.

Now, aob = a + b – ab

= b + a – ba

= boa

So,

aob = boa for all a, b ∈ Q – {-1}

Hence, o is commutative on Q – {-1}

Let us prove the commutativity of *

Let a, b ∈ Q₀

a * b = (3ab/5)

= (3ba/5)

= b * a

So, a * b = b * a, for all a, b ∈ Q₀

Let us prove the associativity of *

Let a, b, c ∈ Q₀

a * (b * c) = a * (3bc/5)

= [a(3bc/5)]/5

= 3 abc/25

(a * b) * c = (3ab/5) * c

= [(3ab/5)c]/5

= 3abc/25

So, a * (b * c) = (a * b) * c, for all a, b, c ∈ Q₀

Thus * is associative on Q₀

Let us find the identity element

Let e be the identity element in Z with respect to *

Such that, a * e = a = e * a ∀ a ∈ Q₀

a * e = a and e * a = a, ∀ a ∈ Q₀

3ae/5 = a and 3ea/5 = a, ∀ a ∈ Q₀

e = 5/3 ∀ a ∈ Q₀ [because a ! = 0]

Hence, 5/3 is the identity element in Q₀ with respect to *.

(i) Let us prove the commutativity of *

Let a, b ∈ Z. Then,

a * b = a + b – 4

= b + a – 4

= b * a

So,

a * b = b * a, ∀ a, b ∈ Z

Thus, * is commutative on Z.

Now, let us prove associativity of Z.

Let a, b, c ∈ Z. Then,

a * (b * c) = a * (b + c – 4)

= a + b + c - 4 – 4

= a + b + c – 8

(a * b) * c = (a + b – 4) * c

= a + b – 4 + c – 4

= a + b + c – 8

So,

a * (b * c) = (a * b) * c, for all a, b, c ∈ Z

Thus, * is associative on Z.

(ii) Let e be the identity element in Z with respect to *

Such that, a * e = a = e * a ∀ a ∈ Z

a * e = a and e * a = a, ∀ a ∈ Z

a + e – 4 = a and e + a – 4 = a, ∀ a ∈ Z

e = 4, ∀ a ∈ Z

Thus, 4 is the identity element in Z with respect to *.

(iii) Let a and b ∈ Z be the inverse of a. Then,

a * b = e = b * a

a * b = e and b * a = e

a + b – 4 = 4 and b + a – 4 = 4

b = 8 – a ∈ Z

So, 8 – a is the inverse of a ∈ Z

Let A be the first term and D be the common difference of A.P.

a_p = a, ∴ A + (p – 1)D = a ….. (1) 

a_q = b, ∴ A + (q – 1)D = b ……. (2) 

a_r = c, ∴ A + (r – 1)D = c …….. (3) 

∴ a (q – r) + b (r – p) + c (p – q) = [A + (p – l) D] (q – r) + [A + (q – 1) D] 

(r – p) + [A + (r – 1) D] (p – q) [Using (1), (2) and (3)] 

= (q – r + r – p + p – q)A + [(p – l)(q – r) + (q – l)(r – p) + (r – l)(p – q)]D

= (0) A + (pq – pr – q + r + qr – pq – r + p + pr – p – qr + q)D 

= (0)A + (0)D = 0.

 \((x^3 - \frac{3}{x^2})^{15}\)

Here x → x³, a = \((\frac{-3}{x^2})^r\) and n = 15 

T_r+1 = ¹⁵C_r.(x³)^{15- r} \((\frac{-3}{x^2})^r\). 

= ¹⁵C_r.x^45-r .x^-2r (-3)^r 

= ¹⁵Cr.(-3)^r .x^45-5r 

To find the term independent of x we have 45 – 5r = 0 

:. 45 = 5r ⇒ r = 9 

T₉₊₁ = ¹⁵C₉.(-3)⁹ .x⁰ 

T₁₀ = ^-15C₉.(3)⁹ is the term independent of x.

Here x → x² a → \(\frac{3a}{x}\), n =15 

∴ T_r+1 = ¹⁵C_r.(x²)^15-r(\(\frac{3a}{x}\))^r 

= ¹⁵C_rx^30-2r.(3a)^rx^r 

= ¹⁵C_r .3^r .a^r .x^30-3r 

To find the coefficient of x¹⁸, we get 30 – 3r = 18 

12 = 3r ⇒ r =4 

∴ T₄₊₁ = ¹⁵C₄.3⁴ .a⁴ .x¹⁸ 

∴ coefficient of x¹⁸ is ¹⁵C₄.(3a)⁴

Here, x → x, a → \(\frac{1}{x^2}\) and n = 17 

T_r+1 = ¹⁷C_r.x^17-r(\(\frac{1}{x^2}\))^r 

= ¹⁷C_rx^17-3r 

To find the coefficient of x⁵, equate the power of x to 5 

∴ 17 – 3r = 5 

12 = 23 ⇒ r = 4 

T₄₊₁ = T₅ = ¹⁷C₄.x⁵ 

∴ The coefficient of x⁵ is ¹⁷C₄

Here, x → x, a → \(\frac{1}{x^2}\)and n = 17 

∴ T_r+1 = ¹⁷C_r.x^17-r(\(\frac{1}{x^2}\))^r 

T_r+1 = ¹⁷C_r.x^17-r-2r 

= ¹⁷C_rx^17-3r 

To find the coefficient of x^-2, equate the power of x to -2 

17 – 3r = -2 ⇒ 17 + 2 = 3r ⇒ 3r ⇒ r = \(\frac{19}{3}\)

Since r is a fraction the coefficient of x^-2 is 0.

To Find : General term, i.e. t_r+1 

For (x² - y)⁶ 

a=x² , b=-y and n=6

General term t_r+1 is given by,

t_r+1 = (_rⁿ) a^n-rb^r

= (⁶_r)(x²)^6-r(-y)^r

Conclusion : General term = (⁶_r)(x²)^6-r(-y)^r

Answer is (A) ⁿC_r x^n-r .a^r

General term in the expansion of (x + a)ⁿ

⇒ T_r+1 = ⁿC_r x^n-r a^r

Given 6 = coefficient of x² in (1 + x)^m

= ^mC₂

∴ 6 = m(m - 1)/(2 x 1)

⇒ m(m - 1) = 12 = 4(3)

∴ m = 4

The general term in the expansion of \((x-\frac{1}{2x})^6\) is

T_{r + 1} = (– 1)^{r 6}C_r x^{6 – r} \((\frac{1}{2x})^r\) = (– 1)^r . ⁶C_r x^{6 – r} . 2 ^{– r} x ^{– r} 

= (– 1)^r . ⁶C_r x^{6 – 2r} . 2^{– r}

Now the power of binomial expansion being 6, (even), the middle term is \((\frac{6}{2}+1)th\) term = 4th term.

∴  T₄ = T₃₊₁ = (-1)³ . ⁶C₃ x^6-6 2^-3 

= \((-1) \times \frac{6\times 5 \times 4}{3 \times 2 \times 1} \times x^0 \times \frac{1}{8} = -\frac{5}{2}\)

Given as

(2x – 1/x²)²⁵

Now, the given expression contains 26 terms.

Therefore, the 11^th term from the end is the (26 − 11 + 1) ^th term from the beginning.

In other words, the 11^th term from the end is the 16^th term from the beginning.
Now,

T₁₆ = T₁₅₊₁ = ²⁵C₁₅ (2x)^25-15 (-1/x²)¹⁵

= ²⁵C₁₅ (2¹⁰) (x)¹⁰ (-1/x³⁰)

= – ²⁵C₁₅ (2¹⁰ / x²⁰)

Then we shall find the 11^th term from the beginning.

T₁₁ = T₁₀₊₁ = ²⁵C₁₀ (2x)^25-10 (-1/x²)¹⁰

= ²⁵C₁₀ (2¹⁵) (x)¹⁵ (1/x²⁰)

= ²⁵C₁₀ (2¹⁵ / x⁵)

Given as

2³ⁿ – 7n – 1

Therefore, 2³ⁿ – 7n – 1 = 8ⁿ – 7n – 1

Now,

8ⁿ – 7n – 1

8ⁿ = 7n + 1

= (1 + 7) ⁿ

= ⁿC₀ + ⁿC₁ (7)¹ + ⁿC₂ (7)² + ⁿC₃ (7)³ + ⁿC₄ (7)² + ⁿC₅ (7)¹ + … + ⁿC_n (7) ⁿ

8ⁿ = 1 + 7n + 49 [ⁿC₂ + ⁿC₃ (7¹) + ⁿC₄ (7²) + … + ⁿC_n (7) ^n-2]

8ⁿ – 1 – 7n = 49 (integer)

Therefore now,

8ⁿ – 1 – 7n is divisible by 49

Or

2³ⁿ – 1 – 7n is divisible by 49.

Thus proved.

Correct option  (a)  4

Explanation:

Coefficient of x¹³ in = (1 – x)⁵ (1 + x + x² + x³) 4 = (1 – x)⁵ ((1 + x)(1 + x²))⁴

= (1 – x) {(1 – x)(1 + x)(1 + x² )}4

= (1 – x) {(1 – x⁴)⁴ 

= (1 – x) (1 – 4x⁴  + 6x⁸ – 4x¹² + x¹⁶) 

∴  coefficient of x¹³ is  –1 x – 4 = 4

(1 + x)ⁿ= ⁿC₀ 1ⁿ + ⁿC₁ 1ⁿ + ⁿC₁ 1^n-1 x¹

ⁿC₂ 1^n-2 x² + ⁿC₃ 1^n-3x³ +….

Putting x = 1

(1 + 1)ⁿ = ⁿC₀ + ⁿC₁ +ⁿC₂ + ⁿC₃ + ….

Putting x = – 1

(1 – 1)ⁿ = ⁿC₀ – ⁿC₁ +ⁿC₂ – ⁿC₃ + ….

Here ⁿC₀ + ⁿC₁ +ⁿC₂ + ⁿC₃ + …. = 2ⁿ … (i)

“C0 – “Ci + ”C2 – “C3 + … = 0 … (ii)

Adding equation (i) and (ii)

2[ⁿC₀ + ⁿC₂ + ⁿC₄ + …] = 2ⁿ

⇒ ⁿC₀ + ⁿC₂ + ⁿC₄ + … = 2^{n – 1}

or C₀ + C₂ + C₄ + … = 2^{n – 1}

(1 + x)ⁿ = ⁿC₀ + ⁿC₁x+ⁿC₂x²+ ⁿC₃x³+ ….

Putting x = 1

(1 + 1)ⁿ = ⁿC₀ + ⁿC₁ +ⁿC₂ + ⁿC₃ + …. +  ⁿC_n

or    2ⁿ = ⁿC₀ + ⁿC₁ +ⁿC₂ + ⁿC₃ + …. + ⁿC_n

Here, putting n = 30.

2³⁰ = 1+ ³⁰C₁ +³⁰C₂ + ³⁰C₃ + …. + ³⁰C₃o

⇒ 1+ ³⁰C₁ +³⁰C₂ + ³⁰C₃ + …. + ³⁰C₃o  = 2³⁰

⇒ ³⁰C₁ + ³⁰C₂ + ³⁰C₃ + … + ³⁰c₃₀ = 2³⁰ – 1

We know that

(1 – x)^-1 = 1 + x + x² + x³ +…           … (i)

and (1 – x)^-3 = 1 + 3x + 6x² + 10x³ + … (ii)

L.H.S. = (1 + x + x² + x³ +…)(1 + 3x + 6x² + …)

= {(1 – x)^-1} {(1 – x)^-3}

= {(1 – x)²}^-2

= {(1 – x)^-2}²

= (1 + 2x + 3x² +…)²

= R.H.S.

By using binomial theorem,

(a + b)⁴ = ⁴C₀ a⁴ b⁰ + ⁴C₁ a³ b¹
+ ⁴C₂ a² b²+ ⁴C₃ ab³ + ⁴C₄ a⁰ b⁴

= ⁴C₀ a⁴ + ⁴C₁ a³ b¹ + ⁴C₂ a² b²
+ ⁴C₃ ab³ + ⁴C₄ b⁴ …. (1)

and (a – b)⁴ = ⁴C₀ a⁴ (-b)⁰+ ⁴C₁ a³ (-b)¹
+ ⁴C₂ a² (-b)² + ⁴C₃ a¹ (-b)³ + ⁴C₄ a⁰ (-b)⁴

= ⁴C₀ a⁴ – ⁴C₁ a³ b + ⁴C₂ a² b²
– ⁴C₃ ab⁴ + ⁴C₄ b⁴ …. (2)

From equation (1) and (2) we have,

(a + b)⁴ – (a – b)⁴

= [⁴C₀ a⁴ + ⁴C₁ a³ b +⁴C₂ a² b² + ⁴C₃ ab³ + ⁴C₄ b⁴]
– [⁴C₀ a⁴ – ⁴C₁ a³ b + ⁴C₂ a² b² –⁴C₃ ab³ + ⁴C₄ b⁴]

= ⁴C₀ a⁴ + ⁴C₁ a³ b + ⁴C₂ a² b² + ⁴C₃ ab³ + ⁴C₄ b⁴
– ⁴C₀ a⁴ + ⁴C₁ a³b – ⁴C₂ a²b² + ⁴C₃ ab³ – ⁴C₄ b⁴

= 2. ⁴C₁ a³ b + 2. ⁴C₃ ab³

= 2ab [⁴C₁ a² + ⁴C₃ b²]

= 2ab [4a² + 4b²]  [∴ ⁴C₁ = 4, ⁴C₃ = 4]

= 2ab (a² + b²)

Hence, (a + b)⁴ – (a – b)⁴ = 8ab (a² + b²)

Now, putting a= √3 and b = √2

(√3 + √2)⁴ – (√3 – √2)⁴

= 8 √3 × √2 [(√3)² + (√2)²]

= 8 √6 (3 + 2)

= 8 √6 × 5 – 40 √6

Hence (√3 + √2 )⁴ – (√3 – √2 )^4 = 40 √6

Correct option (a)  2¹⁹(2²⁰ – 1)

Explanation: 

Put x = 1 and x = –1 and adding we get 4²⁰+2²⁰ = 2(a₀ + a₂ + a₄ +....+ a₃₈ + a₄₀)

⇒ 2³⁹ + 2¹⁹ = a₀ + a₂  + ..... + a₃₈ + 2²⁰ 

∴ a₀ + a₂ + a₄ +.....+ a₃₈ 

=2³⁹ + 2¹⁹ – 2²⁰ 

= 2¹⁹(2²⁰ + 1– 2)

= 2¹⁹(2²⁰ – 1)

Given expansion is :

(1 + x – 2x²)⁶ = 1 + a₁x+ a₂x² + a₃x³ + … + a₁₂x¹² … (i)

Putting x = – 1 in equation (i)

{1 + 1 – 2(1)²}⁶ = 1+ a₁ + a₂ + a₃ + … + a₁₂

⇒ (2 – 2)⁶ = 1 + a₁ + a₂ + a₃ + … + a₁₂

⇒ 1 + a₁ + a₂ + a₃ + … + a₁₂ = 0 … (ii)

Putting x = – 1 in equation (i)

{(1 – 1 – 2(-1)²}⁶ = 1 – a₁ + a₂ + a₃ + … + a₁₂

⇒ (-2)⁶ = 1 – a₁ + a₂ + a₃ + … + a₁₂

⇒ 1- a₁ + a₂ + a₃ + … + a₁₂ = 64       … (iii)

Adding equation (ii) and (iii) we get

⇒ 2 + 2a₂ + 2a₄ + … + a₁₂ = 64

⇒   2(1 + a₂ + a₄ + … + a₁₂₎ = 64

⇒  1 + a₂ + a₄ + … + a₁₂ = 64/2 = 32

⇒ a₂ + a₄ + a₆ … + a₁₂ = 32 – 1 = 31

Hence, a₂ + a₄ + a₆ … + a₁₂ =31.

Hence proved.

We have T_{r + 1} = ^{m + n}C_r a^r

⇒ Coefficient of a^r = ^{m + n}C_r

∴ Coefficient of a^m = ^{m + n}C_m

Similarly, coefficient of aⁿ = ^{m + n}C_n

= ^{m + n}C_{(m + n) - n}

= ^{m + n}C_m

∴ Coefficient of a^m = coefficient of aⁿ

Answer : (a) 16. ⁸C₄   

Sum of the coefficients of the expansion \((\frac{1}{x} +2x)^n\)  _{= 6561}

Putting x = 1, (1 + 2)ⁿ = 6561 ⇒ 3ⁿ = 3⁸ 

⇒ n = 8

 \(\therefore\) T_{r + 1} in the expansion of \((\frac{1}{x} +2x)^8\) 

= ⁸C_r \((\frac{1}{x})^{8-r}\) (2x)^r = ⁸C_r 2^r x^2r-8 

Since this term is independent of x, 2r – 8 = 0 ⇒ r = 4.

∴ Reqd. term = ⁸C₄ . 2⁴ 

= 16 . ⁸C₄.

Number of bacteria at the beginning = 30 

Number of bacteria after 1 hour = 30 × 2 = 60 

Number of bacteria after 2 hours = 30 × 2² = 120 

Number of bacteria after 4 hours = 30 × 24 = 30 × 16 = 480 

∴ Number of bacteria after n^th hour = 30 × 2ⁿ

To get the sum of coefficients of (x + y)⁸ , 

Put x = 1 and y = 1, we get

 (1 + 1)⁸ = 2⁸ = 256

(1, 2)⁴⁰⁰⁰ = (1 + 0.2)⁴⁰⁰⁰ 

Now, expanding by binomial theorem, we get 

(1 + 0.2)⁴⁰⁰⁰ = ⁴⁰⁰⁰ C₀ (1)⁴⁰⁰⁰  (0.2)⁰ + ⁴⁰⁰⁰ C₁ (1^{4000  – 1} (0.2)^{1 +4000} C₃ (1)^{4000 – 3} (0.2)3 + …

 = 1 + 4000(0.2) + other terms of the expansion 

= 1 + 800 + other terms of the expansion 

= 801 + other terms of the expansion 

Hence, (1, 2)⁴⁰⁰⁰  > 800

Let (r + 1)^th term be independent of in the given expression 

Now T_{r + 1} = ¹⁰Cr (3x)^{10 – r} ( –\(\frac{1}{2x^3}\))^r

= ¹⁰Cr (3x)^{10 – r} ( –\(\frac{1}{2}\))^r x^{30 − 6}

This term is independent of x, if 

30 – 6r = 0 

⇒ r = 5 

So, (5 + 1)^th i.e 6th terms is independent of x

Given, 

{(2x + y3 )⁴ }⁷ 

= (2x + y³ ) ²⁸ 

Hence, there are 29 terms in the expansion

Answer : (c)  (3n + 1)th term 

(1 – 3x + 3x² – x³)²ⁿ = ((1 – x)³)²ⁿ = (1 – x) ⁶ⁿ 

^∴ Middle term = \((\frac{6n}{2}+1)\)th term 

= (3n+1)th term.

(c) 26 is the correct choice since the total number of terms are 52 of which 26 terms get cancelled.

The correct option is (C) 26

Explanation:

We have to expand (x + a)⁵¹ – (x – a)⁵¹
At first,  (x + a)⁵¹ = ⁵¹C₀ x⁵¹ + ⁵¹C₁ x^50 . a + ⁵¹C₂ x^49 . a² + ...... + ⁵¹C₅₁ a⁵¹
then, (x – a)⁵¹ = ⁵¹C₀ x⁵¹ - ⁵¹C₁ x^50 . a + ⁵¹C₂ x^49 . a² - ...... - ⁵¹C₅₁ a⁵¹

When we subtract both the values i.e. (x + a)⁵¹ – (x – a)⁵¹ we get, 
2( ⁵¹C₁ x^50 . a + ⁵¹C₃ x^48 . a³ + ...... + ⁵¹C₅₁ a⁵¹)

Thus count the number of terms that is number of odd numbers up to 51. 

i.e 1, 3, 5, 7, ......, 49, 51

Apply AP:
a + (n-1)d = 51
1 + (n-1)2 = 51
n = 26
So, the total numbers of terms in the expansion is 26.