JEE Advanced Level Test: Binomial Theorem- 1 - JEE MCQ

For the sum of coefficient, put x = 1, to obtain the sum 
which implies (1 + 1 - 3)²¹³⁴ = 1

C(n, r)  + c(n -1, r)  +  C(n - 2, r) +  . . .   + C(r, r)
= ^r+1C_r+1  + ^r+1C_r + ^r+2C_r + . . .  .  +  ^n-1C_r + ⁿC_r
= ⁿ⁺¹C_r+1 (applying same rule again and again)        
(∴ ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r)

(x² + (x⁶ - 1)^1/2)⁵ + (x² - (x⁶ - 1)^1/2)⁵
= 2(⁵C₀(x²)⁵ + ⁵C₂ (x²)³ (x⁶ - 1) + ⁵C₄ (x⁶ - 1)²)
Here last term is of 14 degree.

The  general term 


The term independent of x, (or the constant term) corresponds to x^18-3r being x⁰ or 18 - 3r = 0 ⇒ r = 6 .

Hence t₁ < t₂ < t₃ < t₄ < t₅ <t₆ <t₇ < t₈ < t₉ < t₁₀
Hence, t₈ is  the greatest   term and its   value is

(1+ 8)ⁿ⁺¹ = ⁿ⁺¹C₀ + ⁿ⁺¹C₁(8)¹+ ⁿ⁺¹C₂(8)² + ⁿ⁺¹C₃ (8)³ + .... + ⁿ⁺¹C_n+1(8)ⁿ⁺¹     (1)
⇒ 9ⁿ⁺¹ = 1+(n+1)x8+ⁿ⁺¹C₂(8)³ + ....+ⁿ⁺¹C_n-1(8)ⁿ⁺¹
⇒ 9ⁿ⁺¹ - 8n - 9 = (8)² [ⁿ⁺¹C₂ + ⁿ⁺¹C₃(8)+ⁿ⁺¹C₄(8)²+...+ⁿ⁺¹C_n+1(8)^n-1]
⇒ 9ⁿ⁺¹ - 8n - 9 = 64 x an integer ⇒ 9ⁿ⁺¹ - 8n - 9 is divisible by 64.

∴ T_r+1 = ⁹C_r (3^1/2)^9-r (2^1/3)^r = ⁹C_r 3^(9-r)/2.2^r/3
For first integral term for r = 3;
T₃₊₁ = ⁹C₃3³.2¹ i.e., T₃₊₁ = T₄ (4^th term)

(2^1/5 + 3^1/10)⁵⁵ 
Total term = 55 + 1 = 56
T_r+1 = ⁵⁵C_r 2^(55-r)/5 3^r/10 
Here r = 0, 10, 20, 30, 40, 50
Number of rational terms = 6;    
Number of rational terms = 56 - 6 = 50

We have, (2 + 3 - 4z)ⁿ = {2 + (3 - 4)}ⁿ
= ⁿC₀ (2x)ⁿ (3y - 4z)° + ⁿC₁ (2x)^n-1(3y-4z)¹ + ⁿC₂ (2x)^n-2 (3y - 4z)² +.. + ⁿC_n-1(2x)¹ (3y - 4z)^n-1 + ⁿC₂(3y-4z)n
Clearly, the first term in the above   expansion gives one term, second term gives two terms, third term gives three terms and so on.
So, Total number of term = 1 +2+3+...+n+(n+1) = 

Given expansion is 
∴ General term = T_r+1 = ¹²C_r (a/x)^12-r (bx)^r = ¹²C_r (a)^12-r b^r x^-12+2r
Since, we have to find coefficient of x-¹⁰
∴ -12 + 2r = -10 ⇒ r = 1
Now, then  coefficient of x^-10 is ¹²C₁ (a)¹¹ (b)¹ = 12a¹¹b

∴ n = 4,a = 2

Correct Answer :- A

Explanation : (x + 1/x)¹¹, it has 12 terms

Therefore, there are two middle terms (6th and 7th)

T₆T₇ = [¹¹C₅(x)⁶(1/x)⁵] * [¹¹C₆(x)⁵(1/x)⁶]

¹¹C₅ ¹¹C₆

Here 2n is even   integer, therefore,

term  will be the middle term.
Now, ( n + 1)^th term in (1-x)²ⁿ = ²ⁿC_n(1)^2n-n(-x)ⁿ = ²ⁿC_n(-x)ⁿ

Given  sum = sum of odd terms 

(1- x+2x²)¹⁰ = a₀+a₁x+a₂x² +..a₁₉x¹⁹ + a₂₀x²⁰ 
Replacing x by-x, (1+x+ 2x²)¹⁰ = a₀ - ₁x+a₂x²...a₁₉x¹⁹ +a₂₀x²⁰) 
Subtracting and putting x = 1, 2¹⁰ - 4¹⁰ = 2(a₁ +a₃ +...+a₁₉)
⇒ 

Here 2n + 2 is even  
Greatest  coefficient

(1+x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² - ..+ ⁿC_nxⁿ....       (i)
(x+1)ⁿ = ⁿC₀xⁿ + ⁿC₁x^n-1 + ⁿC₂x^n-2 + ..... +ⁿC_n ......(ii)

Multiply (i) and (ii)  and consider  the coefficient xⁿ of both sides, we have 
 coefficient of 

We know (1+x)ⁿ = C₀ +C₁x+C₂x² +C₃x³ + ...+ C_nxⁿn(1+ x)^r-1 = C₁+ 2C₂x + 3C₃x² +4C₄x³ +...+nC_nx^n-1 
Putting x =1 and —1, then add
n2^n-1 = C₁ + 2C₂ +3C₃ +4C₄ +..+nC_n
0= C₁ - 2C₂ + 3C₃ - 4C₄ +... ⇒ n2^n-2
C₁ + 3C₃ +5C₅ +...

= 8(1+31)¹⁵ = 8{¹⁵C₀ + ¹⁵C₁31+..+¹⁵C₁₅(31)¹⁵}
2⁷⁸ = 8 + an integer multiple of 31;

Given expression

Given expansion  



Number of dissimilar  term = (2n + 1) + (2n -1) + ....(2n - 3) + .... + 5 + 3 + 1

Where ao = sum of all absolute terms   = 1 +¹⁰⁰C₂ .2 + ....
Similarly  a₁ , a₂ , ...a₁₀₀ and  b₁ , b₂ ...b₁₀₀ are coefficients  obtained   after simplification. 
∴ Total  number  of terms  = 1 + 100 + 100 = 201

Subtract  above  equations, 

 

⇒ 


Coefficient of x⁵⁰ in S = coefficient of x⁵⁰ in

Given expression 




⇒ 




which is independent of x if 

Hence  required  coefficient

where I is integer and 0 < f <1.

Now, 


∴ I + f + F is integer.

Given m - n = 3


⇒ 
⇒ (m - n)² - m + n = -12 ⇒ m + n = 9 + 12 = 21   (2) using (1)
Solving  (1) and (2), we get m = 12.

t²⁴ in (1+t²)¹²(1+t¹²)(1+t²⁴) is
Here, coefficient of t²⁴ in 

⇒ coefficient of t²⁴ in

⇒ coefficient of t²⁴ in 


coefficient of 
 coefficient of