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

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: 

\(\displaystyle \sum_{\mathrm{k}=0}^{6}{ }^{51-\mathrm{k}} \mathrm{C}_{3}\)

= 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 \( T_r = C(n, r) \cdot a^{n-r} \cdot b^r \).
• Where \( C(n, r) \) is the binomial coefficient, representing the number of ways to choose \( r \) elements from \( n \) elements.
• The largest term in a binomial expansion occurs when \( r = \left( \frac{n}{2} \right) \) (approximately), and we find the term corresponding to this value of \( r \).
• For any real number \( a_0, a_1, \ldots, a_{23} \), the largest term occurs at a specific \( r \) value that maximizes the binomial coefficient.

Calculation:

We are given the following equation:

• \( \left( 1 + \frac{2}{x} \right)^{23} = \sum_{r=0}^{23} a_r x^r \)
• We need to find the value of \( r \) where \( a_r \) is the largest term.

By comparing the binomial expansion of \( \left( 1 + \frac{2}{x} \right)^{23} \), we can determine the value of \( r \) where the coefficient \( a_r \) is maximized.

First, calculate the general term of the binomial expansion:

\( T_r = C(23, r) \left( \frac{2}{x} \right)^r \)

The ratio of successive terms \( \frac{T_{r+1}}{T_r} \) helps determine the maximum term:

\( \frac{T_{r+1}}{T_r} = \frac{C(23, r+1) \left( \frac{2}{x} \right)^{r+1}}{C(23, r) \left( \frac{2}{x} \right)^r} \)

Simplifying the ratio gives:

\( \frac{T_{r+1}}{T_r} = \frac{23 - r}{r+1} \cdot \frac{2}{x} \)

Set \( \frac{T_{r+1}}{T_r} = 1 \) to find the value of \( r \):

\( \frac{23 - r}{r + 1} \cdot \frac{2}{x} = 1 \)

Solve for \( r \):

\( r = 6 \)

Conclusion:

Hence, the value of r that maximizes the term is \( r = 6 \).

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: \(\rm \left(2 x+\frac{1}{x^{7}}+3 x^{2}\right)^{5}\) 

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.

Calculation: 

\(S = (2 + √{3})^8\)

For sum of rational terms

⇒ ⁸C₀(2)⁸ + ⁸C₂ (2)⁶ (√3)² + ⁸C₄ 2⁴ (√3)⁴ + ⁸C₆ (2)⁶ (√3)⁶ + ⁸C₈(2)⁸ (√3)⁸

= 1. 256 + 28. 64. 3 + 70. 16. 9 + 28. 4. 27 + 1. 81

= 256 + 5376 + 10080 + 3024 + 81

= 18817.

Hence, the Correct answer is Option 4.

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 = \(a (r^n-1)\over(r- 1)\); where r >1

Sum of n terms = s = \(a (1- r^n)\over(1- r)\); where r <1

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 = \(\rm 2^n - 1\over 2-1\) = 1 × \(\rm 2^n - 1\over 2-1\)

Comparing it with a G.P sum = a × \(\rm r^n - 1\over r-1\), we get a = 1 and r = 2

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

Concept:

\(\rm ^n C_r = {n!\over(r!(n - r)!)}\)

(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:

Expansion of (1 + x)n:

\(\rm (1+x)^n = 1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3 +....\)

Calculation:

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

\(\rm (1+x)^m= 1+mx+\frac{m(m-1)}{2!}x^2+\frac{m(m-1)(m-2)}{3!}x^3 +....\)

So, the third term in the binomial expansion of (1 + x)m is \(\rm \frac{m(m-1)}{2!}x^2\)

\(\rm \frac{m(m-1)}{2!}x^2\) = (-1/8)x²

⇒ \(\rm \frac{m(m-1)}{2}= \frac {-1}{8}\)

⇒ 4m² - 4m + 1 = 0

⇒ (2m - 1)² = 0

⇒ 2m - 1 = 0

∴ m = \(\frac 12\)

Concept:

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

\(\rm {T_{\left( {r\; + \;1} \right)}} = \;{\;^n}{C_r} \times {x^{n - r}} \times {y^r}\)

Expansion of (1 + x)ⁿ:

\(\rm (1+x)^n = 1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3 +....\)

Calculation:

To Find: coefficient of x² in the expansion of \(\rm (4-5x^2)^{-1/2}\)

\(\rm (4-5x^2)^{-1/2} = 4^{-1/2}\left(1-\frac{5}{4}x^2 \right )^{-1/2}\\ \text{As we know}\;\rm (1+x)^n = 1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3 +....\\\therefore 4^{-1/2}\left(1-\frac{5}{4}x^2 \right )^{-1/2} = 2^{-1}\left[1 + \left(-\frac{5}{4}x^2 \right ) \times (\frac{-1}{2}) + ... \right ]\)

Now, the coefficient of x² in the expansion = \(2^{-1} \times \frac{-5}{4} \times \frac{-1}{2} = \frac{5}{16}\)

Concept:

Binomial expansion of (x + y)ⁿ is given by

(x + y)n = ⁿC₀xⁿ + nC1xn-1y + nC2xn-2y²+.....+ nCn-1xy^n-1 + nCnyn  

Calculation:

Given,  \(\displaystyle\sum_{r=0}^n\)2r C(n, r)

Expanding the expression,

⇒ nC020 + nC1 2¹ + nC222+.....+ nCn-12n-1 + nCn2n  

⇒ nC01n 20 + nC1 1n-1 2¹ + nC21n-2 22+.....+ nCn-1 2n-1 + nCn2n  

Comparing with binomial expansion x = 1 and y = 2

⇒  \(\displaystyle\sum_{r=0}^n\)2r C(n, r) = (1 + 2)ⁿ

⇒  \(\displaystyle\sum_{r=0}^n\)2r C(n, r) = 3n

∴ The correct option is (2).

Concept:

• Number of terms in the expansion of (x₁ + x₂ + x₃ +......+ x_r)ⁿ is =  \(^{n+r-1}C_{n}\), Where n = exponent of the term to be expanded, r = number of terms to be expanded
• ​ \(^{n}C_{r}=\frac{n!}{(n-r)!~r!}\)

Explanation:

Number of terms in (x + y + z)10 = \(^{10+3-1}C_{10}\) (Since, n = 10 and r = 3)

⇒ \(^{12}C_{10}\)

Since, \(^{12}C_{10}=\frac{12!}{(12-10)!~10!}\)

\( \Rightarrow ^{12}C_{10}=66\)

Formula used:

\(^nC_r=\frac{n!}{r!(n-r)!}\)

n! = n × (n - 1) × (n - 2).......3 × 2 × 1 

Calculation:

Given that,

4 × nC5 = 9 × n-1C5

Using the above formula

\(4×\frac{n!}{5!(n-5)!}=9×\frac{(n-1)!}{5![(n-1)-5]!}\)

⇒ \(4×\frac{n× (n-1)!}{5!(n-5)(n-6)!}=9×\frac{(n-1)!}{5![(n-6]!}\)

⇒ \(\frac{4n}{(n-5)} = 9\)

⇒ 9n - 45 = 49

⇒ 5n = 4n

⇒ n = 9

Concept:

We have (x + y) n = nC0 xn + nC1 xn-1 . y + nC2 xn-2. y2 + …. + nCn yn

• General term: General term in the expansion of (x + y) n is given by
• \({{\rm{T}}_{\left( {{\rm{r\;}} + {\rm{\;}}1} \right)}} = {\rm{\;}}{{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}} \times {{\rm{x}}^{{\rm{n}} - {\rm{r}}}} \times {{\rm{y}}^{\rm{r}}}\)
• In the binomial expansion of (x + y)n , the rth term from end is (n – r + 2)th term.

Note: 

• In the binomial expansion of (x + y)n, the rth term from end = In the binomial expansion of (y + x)n, the rth term from the start.
• If we interchange the term x → y, it will give rth term from the beginning.

Calculation:

We have to find 9th term from the end in (x – 1/x) 12

We know that rth term from end means (n – r + 2)th term from start.

So. 9th term from the end = [12 – 9 + 2]th term from start = 5th term from start

General term: \({{\rm{T}}_{\left( {{\rm{r\;}} + {\rm{\;}}1} \right)}} = {\rm{\;}}{{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}} \times {{\rm{x}}^{{\rm{n}} - {\rm{r}}}} \times {{\rm{y}}^{\rm{r}}}\)

\(\Rightarrow {{\rm{T}}_5} = {\rm{\;}}{{\rm{T}}_{\left( {4 + {\rm{\;}}1} \right)}} = {\rm{\;}}{{\rm{\;}}^{12}}{{\rm{C}}_4} \times {{\rm{x}}^{12 - 4}} \times {\left( {\frac{{ - 1}}{{\rm{x}}}} \right)^4}\)

\( = {{\rm{\;}}^{12}}{{\rm{C}}_4} \times {{\rm{x}}^8} \times \frac{1}{{{{\rm{x}}^4}}}\)

= ¹²C₄ x⁴

Concept:

The expansion of (x + y)ⁿ is given by the binomial expansion. The expansion will be 

(x + y)ⁿ = ⁿC₀ xⁿ + ⁿC₁ x^n-1 y + ⁿC₂ x^n-2 y² + ... + ⁿC_r x^n-r y^r + ... + ⁿC_n yⁿ;

The general term is given by 

T_r =  nCr xn-r yr ;

Calculation:

Given expansion is \(\left(3x^2 + \frac 1 x\right)^{30}\)

Let the r^th term has x¹² and the coefficient be A;

⇒ T_r = A x¹²;

The general r^th term of the given expansion will be 

T_r = \(^{30}C_r (3x^2)^{30-r}(\frac {1}{x})^{r} \)

⇒ T_r = ³⁰C_r 3^30-r × x^60-2r × x^-r = 30Cr 330-r × x60-3r

Equating the power of x with 12,

⇒ 60 - 3r = 12 ⇒ r = 16

Now the coefficient will be 

A = 30Cr 330-r = ³⁰C₁₆ 3¹⁴;

Concept:

Binomial Expansion:

(a + b)ⁿ = C₀ aⁿ b⁰ + C₁ a^n-1 b¹ + C₂ a^n-2 b² + … + C_r a^n-r b^r + … + C_n-1 a¹ b^n-1 + C_n a⁰ bⁿ, where C₀, C_1, …, C_n are the Binomial Coefficients defined as C_r = ⁿC_r = \(\rm \frac{n!}{r!(n-r)!}\).

(1 - x)^-n = 1 + ⁿC₁ x + ⁿ⁺¹C₂ x² + ⁿ⁺²C₃ x³ + ....
Algebraic Identities:

a³ - b³ = (a - b)(a² + ab + b²).
Calculation:

We note that 1³ - x³ = (1 - x)(1 + x + x²).

⇒ 1 + x + x2 = \(\rm{1-x^3\over1-x}\).

Now, (1 + x + x2)-3 can be written as:

= \(\rm\left({1-x^3\over1-x}\right)^{-3}\)

= (1 - x)3 × (1 - x³)^-3

= (1 - 3x + 3x² - x³)(1 + ³C₁ x³ + ⁴C₂ x⁶ + ... higher powers of x)

x⁶ will be obtained by multiplying (1 and ⁴C₂ x⁶) and (-x³ and ³C₁ x³)

∴ Coefficient of x⁶ will be: (1 × 4C2) + (-1 × 3C1) = 6 - 3 = 3.