Binomial Theorem MCQ Quiz - Objective Question with Answer for Binomial Theorem - Download Free PDF

Concept:

Sum of Binomial Coefficients and Greatest Binomial Coefficient:

• The sum of binomial coefficients in the expansion of is calculated by substituting x = 1 and y = 1. The result is .
• To find the greatest binomial coefficient, we analyze the coefficients where r is the term index in the expansion. The greatest coefficient occurs near the middle term(s).
• Key Formulae:
  • Sum of binomial coefficients:
  • Binomial coefficient:
  • Greatest binomial coefficient: For even n, it occurs at r = n/2. For odd n, it occurs at r = (n-1)/2 and r = (n+1)/2.

Calculation:

Given,

Sum of binomial coefficients =

We calculate n:

⇒

Greatest Binomial Coefficient:

For (even), the greatest binomial coefficient occurs at .

⇒ The term index is r = 4, which corresponds to the 5th term (since indexing starts from 0).

∴ The greatest binomial coefficient occurs in the 5th term.

Concept:

A term in the binomial expansion of (a + b)ⁿ is given by T_k+1 = C(n, k) × a^n-k × b^k.

For a term to be rational, the exponents of both √3 and 5^1/4 must be integers.

Formula Used:

In (√3)^n-k, n-k must be even for it to be rational.

In (5^1/4)^k, k must be a multiple of 4 for it to be rational.

Calculation:

Let n = 12:

⇒ For √3^n-k to be rational, n-k must be even.

⇒ Since n = 12, k must also be even.

⇒ For (5^1/4)^k to be rational, k must be a multiple of 4.

⇒ The values of k that satisfy both conditions (k is even and a multiple of 4) are:

⇒ k = 0, 4, 8, and 12.

⇒ These correspond to 4 rational terms in the expansion.

Hence, the Correct answer is Option 3. 

Calculation: 

= ⁵¹C₃ + ⁵⁰C₃ + ⁴⁹C₃ +.....+ ⁴⁵C₃

= ⁴⁵C₃ + ⁴⁶C₃ +.....+ ⁵¹C₃

= ⁴⁵C₄ + ⁴⁵C₃ + ⁴⁶C₃ +.....+ ⁵¹C₃ - ⁴⁵C₄

= (ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r)

= ⁵²C₄ - ⁴⁵C₄

Hence, the correct answer is Option 3.

Concept:

Binomial Expansion and Maximum Term:

• In a binomial expansion, the general term can be expressed as .
• Where is the binomial coefficient, representing the number of ways to choose elements from elements.
• The largest term in a binomial expansion occurs when (approximately), and we find the term corresponding to this value of .
• For any real number , the largest term occurs at a specific value that maximizes the binomial coefficient.

Calculation:

We are given the following equation:

• We need to find the value of where is the largest term.

By comparing the binomial expansion of , we can determine the value of where the coefficient is maximized.

First, calculate the general term of the binomial expansion:

The ratio of successive terms helps determine the maximum term:

Simplifying the ratio gives:

Set to find the value of :

Solve for :

Conclusion:

Hence, the value of r that maximizes the term is .

Concept:

• Multinomial Expansion: For an expression of the form (a + b + c)ⁿ, each term in the expansion is of the form:  (n! / (r₁! r₂! r₃!)) × a^r₁ × b^r₂ × c^r₃ where r₁ + r₂ + r₃ = n.
• Constant Term: A term with x⁰ (i.e. no x) is called the constant term.
• We apply the multinomial theorem to find the combination of powers that results in an overall exponent of x equal to zero.

Calculation:

We are given:  

Let general term be: (5! / (r₁! r₂! r₃!)) × (2x)^r₁ × (1/x⁷)^r₂ × (3x²)^r₃

Where r₁ + r₂ + r₃ = 5

Total power of x = r₁ × 1 − 7r₂ + 2r₃

We want constant term

⇒ net power of x = 0

So, r₁ − 7r₂ + 2r₃ = 0 ...(i)

And r₁ + r₂ + r₃ = 5 ...(ii)

Solve the two equations:

From (ii): r₃ = 5 − r₁ − r₂

Sub into (i):

r₁ − 7r₂ + 2(5 − r₁ − r₂) = 0

⇒ r₁ − 7r₂ + 10 − 2r₁ − 2r₂ = 0

⇒ −r₁ − 9r₂ + 10 = 0

⇒ r₁ = 10 − 9r₂

Try integer values of r₂ such that r₁ and r₃ are also integers ≥ 0

If r₂ = 1 ⇒ r₁ = 1, r₃ = 5 − 1 − 1 = 3 

Now compute the coefficient:

Term = 5! / (1! × 1! × 3!) × (2x)¹ × (1/x⁷)¹ × (3x²)³

= 120 / (1 × 1 × 6) × 2x × 1/x⁷ × 27x⁶

= 20 × 2 × 27 = 1080

∴ The constant term in the expansion is 1080.

Concept:

General term: General term in the expansion of (x + y)n is given by

Middle terms: The middle terms is the expansion of (x + y) n depends upon the value of n.

• If n is even, then total number of terms in the expansion of (x + y) n is n +1. So there is only one middle term i.e. \(\rm \left( {\frac{n}{2} + 1} \right){{\rm{\;}}^{th}}\) term is the middle term.
• If n is odd, then total number of terms in the expansion of (x + y) n is n + 1. So there are two middle terms i.e. and  are two middle terms.

Here, we have to find the middle terms in the expansion of 

Here n = 8 (n is even number)

∴ Middle term = 

T5 = T (4 + 1) = 8C4 × (2x) (8 - 4) × 

T5 =  8C₄ × 2⁴

Concept:

(1 + x)n = nC0 × 1(n-0) × x 0+ nC1 × 1(n-1) × x 1 + nC2  × 1(n-2) × x2 + …. + nCn  × 1(n-n) × xn

n^th  term of the G.P. is an = arⁿ⁻¹

Sum of n terms = s = ; where r >1

Sum of n terms = s = ; where r

Calculation:

C(n, 1) + C(n, 2) + _ _ _ _  _ + C(n, n) 

⇒ ⁿC₁ + ⁿC₂ + ... + ⁿC_n 

⇒ ⁿC₀ + nC1 + nC2 + ... + nCn - ⁿC₀

⇒ (1 + 1)ⁿ - ⁿC_o 

⇒ 2ⁿ - 1 =  = 1 × 

Comparing it with a G.P sum = a × , we get a = 1 and r = 2

∴ 2n - 1 = 1 + 2 + 2² + ... +2^n-1 which will give us n terms in total.

Concept:

(1 + x)ⁿ = ⁿC₀ × 1^(n-0) × x ⁰+ ⁿC₁ × 1^(n-1) × x ¹ + ⁿC₂  × 1(n-2) × x² + …. + ⁿC_n  × 1(n-n) × xⁿ

Calculation:

Given expansion is (1 + x)²ⁿ

⇒ ²ⁿC₀ ×1^(2n-0) × x⁰ +  2nC₁ ×1(2n-1) × x¹ + ... +  2nC_2n ×1(2n-2n) × x²ⁿ

First term = 2nC0 ×1 × 1 = 1

Last term =  2nC_2n ×1 × x²ⁿ = 1 × x²ⁿ = x²ⁿ

⇒ Sum = 1 + x²ⁿ

Coefficient of 1 = 1, coefficient of x²ⁿ = 1

∴ sum of the coefficients = 1 + 1 = 2.

CONCEPT:

In the expansion of (a + b)n the general term  is given by: Tr + 1 = nCr ⋅ an – r ⋅ br

Note: In the expansion of (a + b)n , the rth term from the end is [(n + 1) – r + 1] = (n – r + 2)th term from the beginning.

In the expansion of (a + b)n , the middle term is  term if n is even.

In the expansion of (a + b)n , if n is odd then there are two middle terms which are given by:. 

CALCULATION:

Given: (x + 3)6 

Here, n = 6

∵ n = 6 and it as even number.

As we know that, in the expansion of (a + b)n the middle term is  term if n is even.

Concept:

General term: General term in the expansion of (x + y)n is given by

Middle terms: The middle terms is the expansion of (x + y) n depends upon the value of n.

• If n is even, then total number of terms in the expansion of (x + y) n is n +1. So there is only one middle term i.e. \(\rm \left( {\frac{n}{2} + 1} \right){{\rm{\;}}^{th}}\) term is the middle term.
• If n is odd, then total number of terms in the expansion of (x + y) n is n + 1. So there are two middle terms i.e. and  are two middle terms.

Calculation:

Here, we have to find the middle terms in the expansion of 

Here n = 5 (n is odd number)

∴ Middle term =  and  = 3^rd and 4^th

T3 = T (2 + 1) = 5C2 × (2x) (5 - 2) ×   and T4 = T (3 + 1) = 5C3 × (2x) (5 - 3) ×  

T3 =  5C2 × (2³x) and T4 = 5C3 × 2² × 

T3 = 80x and T4 = 

Hence the middle term of expansion is 80x and 

Concept:

Expansion of (1 + x)n:

Calculation:

Given: the third term in the binomial expansion of (1 + x)m is (-1/8)x²

So, the third term in the binomial expansion of (1 + x)m is 

 = (-1/8)x²

⇒ 

⇒ 4m² - 4m + 1 = 0

⇒ (2m - 1)² = 0

⇒ 2m - 1 = 0

∴ m = 

Concept:

General term: General term in the expansion of (x + y) ⁿ is given by

Calculation:

Given expansion is

General term =  

For the term independent of x the power of x should be zero 

i.e 

⇒ r = 2

Concept:

General term: General term in the expansion of (x + y)ⁿ is given by

Middle terms: The middle terms is the expansion of (x + y) ⁿ depends upon the value of n.

• If n is even, then the total number of terms in the expansion of (x + y) ⁿ is n +1. So there is only one middle term i.e.  term is the middle term.

• If n is odd, then the total number of terms in the expansion of (x + y) ⁿ is n + 1. So there are two middle terms i.e. and  are two middle terms.

Calculation:

Here, we have to find the coefficient of the middle term in the binomial expansion of (2 + 3x) ⁴

Here n = 4 (n is even number)

∴ Middle term =

T₃ = T _{(2 + 1)} = ⁴C₂ × (2) ^{(4 - 2)} × (3x) ²

T₃ = 6 × 4 × 9x² = 216 x²

∴ Coefficient of the middle term = 216

Concept:

General term: General term in the expansion of (x + y)ⁿ is given by

Expansion of (1 + x)ⁿ:

Calculation:

To Find: coefficient of x² in the expansion of 

Now, the coefficient of x² in the expansion = 

Formula used:

(1 + x)n  = [nC0 + nC1 x + nC2 x2 + … +nCn xn]

• C₀ + C₁ + C₂ + … + C_n = 2ⁿ
• C₀ + C₂ + C₄ + … =  2n-1
• C₁ + C₃ + C₅ + … = 2^n-1

Calculation:

(1 + x)⁵⁰  = [⁵⁰C0 + ⁵⁰C1 x + ⁵⁰C2 x2 + … +⁵⁰Cn x⁵⁰]    ----(1)

Here, n = 50

Using the above formula, the sum of odd terms of the coefficient is

S = (50C1 + 50C3­ + 50C5 + ……. + 50C49)

⇒ S = 2^50-1 = 2⁴⁹

∴ Sum of odd terms of the coefficient = 249