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

Concept:

The general term of the binomial (a + b)n is given by T_r+1 = ⁿC_ra^n-rb^r.

Calculation:

Given, \(\rm \left(\frac{\sqrt[5]{3}}{x}+\frac{2x}{\sqrt[3]{5}}\right)^{12}, x\ne 0\)

∴ General term, \(\mathrm{T}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{\mathrm{r}}\left(\frac{3^{1 / 5}}{\mathrm{x}}\right)^{12-\mathrm{r}}\left(\frac{2 \mathrm{x}}{5^{1 / 3}}\right)^r\)

⇒ \(\mathrm{T}_{\mathrm{r}+1}=\frac{{ }^{12} \mathrm{C}_{\mathrm{r}}(3)^{\frac{12-\mathrm{r}}{5}}(2)^{\mathrm{r}}(\mathrm{x})^{2 \mathrm{r}-12}}{(5)^{\mathrm{r} / 3}}\)

For constant term, power of x = 0

⇒ 2r - 12 = 0

⇒ r = 6

∴ \(\mathrm{T}_7=\frac{{ }^{12} \mathrm{C}_6(3)^{6 / 5}(2)^6}{5^2}=\left(\frac{9 × 11 × 7}{25}\right) 2^8 \cdot 3^{1 / 5}\) = α × 28 × \(\rm \sqrt[5]{3}\) 

⇒ α = \(\left(\frac{9 × 11 × 7}{25}\right)\)

⇒ 25α = 9 × 11 × 7 = 693.

∴ The value of 25α is equal to 693.

The correct value is Option 3.

Explanation:

Given:

⇒ Mean x = np = 6....(i)

⇒ SD = \(\sqrt{npq} = \sqrt2\)...(ii)

Dividing (ii) by (i), we get

⇒ \(\frac{npq}{np} = \frac{2}{6}\)

⇒ q= 1/3

Now

⇒ p = 1 – q = 1- 1\3 = 2/3

From (i)

⇒\(n\times \frac{2}{3} = 6\)

⇒n =9

∴ Option (d) is correct.

Explanation:

Given:

(8 + 3√7) 20  = U + V...(i)

(8 - 3√7) 20  = W...(ii)

⇒ (U + V)W =] (8 + 3√7) 20][(8 - 3√7) 20]

= (64 – 63)²⁰ = 1²⁰ = 1

∴ Option (b) is correct.

Explanation:

Given:

(8 + 3√7) 20  = U + V...(i)

(8 - 3√7) 20  = W...(ii)

Here, 0 < W < 1 

Now, adding Eqs. (i) and (ii), we get

U + V +W = (8 + 3√7) 20 + (8 - 3√7) 20

= 2[²⁰C₀8²⁰+ ²⁰C₂8¹⁸. (3√7 )²+........+(3√7)²⁰

⇒ It is an even number.

Also, 0 < V < 1, 0 < W < 1 and U is an integer

Thus,  V + W is an integer

V + W = 1

∴ Option (d) is correct.

Explanation:

Let p(n) = 7ⁿ – 6n

= (6+1)ⁿ – 6n

= 1+6n +ⁿC₂6² + ⁿC₃6³+.......... 6ⁿ -6n)

= 1+ 6²(nC2+ nC3 × 6 +....... 6^n-2)

= 1+ 36(¹⁰⁰C₂ + 100C3 × 6 +........6⁹⁸).....(n =100)

= 1 + 36 k

where k = (100C2 + 100C3 × 6 +........698)

Thus When 7ⁿ – 6n is divided by 36, the remainder will be 1

∴ Option (b) is correct

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.