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 $(x+y{)}^n$ is calculated by substituting x = 1 and y = 1. The result is $2^n$.
• To find the greatest binomial coefficient, we analyze the coefficients $C(n,r)$ where r is the term index in the expansion. The greatest coefficient occurs near the middle term(s).
• Key Formulae:
  • Sum of binomial coefficients: $\text{Sum}=2^n$
  • Binomial coefficient: $C(n,r)=\frac{n!}{r!(n-r)!}$
  • 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 = $2^n=256$

We calculate n:

$2^n=256$

⇒ $2^8=256$

Greatest Binomial Coefficient:

For $n=8$ (even), the greatest binomial coefficient occurs at $r=n/2=8/2=4$.

⇒ 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. 

Concept:

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

Calculation:

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

∴ 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

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

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

⇒ 25α = 9 × 11 × 7 = 693.

∴ The value of 25α is equal to 693.

The correct value is Option 3.

Concept:

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

• $T_{\left(r+1\right)}={}^n{C}_r\times x^{n-r}\times y^r$
• ${}^n{C}_r=\frac{n!}{r!\left(n-r\right)!}$

Calculation:

We have to find coefficient of x³ in ${\left(2x-\frac{3}{x^2}\right)}^9$

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

${\mathrm{T}}_{\left(\mathrm{r}+1\right)}={}^9{\mathrm{C}}_{\mathrm{r}}\times {\left(2\mathrm{x}\right)}^{9-\mathrm{r}}\times {\left(-\frac{3}{x^2}\right)}^{\mathrm{r}}$

$={}^9{\mathrm{C}}_{\mathrm{r}}\times {\left(2\right)}^{9-\mathrm{r}}\times {\left(\mathrm{x}\right)}^{9-\mathrm{r}}\times {\left(-3\right)}^{\mathrm{r}}\times {\left(x^{-2}\right)}^r$

$={}^9{\mathrm{C}}_{\mathrm{r}}\times {\left(2\right)}^{9-\mathrm{r}}\times {\left(-3\right)}^{\mathrm{r}}\times {\left(\mathrm{x}\right)}^{9-\mathrm{r}}\times x^{-2r}$

$={}^9{\mathrm{C}}_{\mathrm{r}}\times {\left(2\right)}^{9-\mathrm{r}}\times {\left(-3\right)}^{\mathrm{r}}\times {\left(\mathrm{x}\right)}^{9-3\mathrm{r}}$

For coefficient of x³;

⇒ 9 – 3r = 3

⇒ 6 = 3r

∴ r = 2

Now, Coefficient of x³ $={}^9{\mathrm{C}}_{\mathrm{r}}\times {\left(2\right)}^{9-\mathrm{r}}\times {\left(-3\right)}^{\mathrm{r}}=={}^9{\mathrm{C}}_2\times {\left(2\right)}^{9-2}\times {\left(-3\right)}^2$

$=\frac{9!}{2!\left(9-2\right)!}\times 2^7\times 3^2=\frac{9\times 8\times 7!}{2\times 7!}\times 2^7\times 3^2$

= 3² × 2² × 2⁷ × 3² = 2⁹ × 3⁴

∴ Option 3 is correct.

Explanation:

Given:

⇒ Mean x = np = 6....(i)

⇒ SD = $\sqrt{npq}=\sqrt{2}$...(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.

Concept:

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

${\mathrm{T}}_{\left(\mathrm{r}+1\right)}={}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}\times {\mathrm{x}}^{\mathrm{n}-\mathrm{r}}\times {\mathrm{y}}^{\mathrm{r}}$

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. $\left(\frac{\mathrm{n}}{2}+1\right){}^{\mathrm{t}\mathrm{h}}$ 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. ${\left(\frac{\mathrm{n}+1}{2}\right)}^{\mathrm{t}\mathrm{h}}$and ${\left(\frac{\mathrm{n}+3}{2}\right)}^{\mathrm{t}\mathrm{h}}$ are two middle terms.

Here, we have to find the middle terms in the expansion of ${(2\mathrm{x}+\frac{1}{\mathrm{x}})}^8$

Here n = 8 (n is even number)

∴ Middle term = $\left(\frac{\mathrm{n}}{2}+1\right)=\left(\frac{8}{2}+1\right)=5\mathrm{t}\mathrm{h}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}$

T5 = T (4 + 1) = 8C4 × (2x) (8 - 4) × ${\left(\frac{1}{\mathrm{x}}\right)}^4$

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 = $\frac{a(r^n-1)}{(r-1)}$; where r >1

Sum of n terms = s = $\frac{a(1-r^n)}{(1-r)}$; 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 = $\frac{2^{\mathrm{n}}-1}{2-1}$ = 1 × $\frac{2^{\mathrm{n}}-1}{2-1}$

Comparing it with a G.P sum = a × $\frac{{\mathrm{r}}^{\mathrm{n}}-1}{\mathrm{r}-1}$, we get a = 1 and r = 2

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

Concept:

${}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}=\frac{\mathrm{n}!}{(\mathrm{r}!(\mathrm{n}-\mathrm{r})!)}$

(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:

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 $\left(\frac{n}{2}+1\right)th$ 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:$\left(\frac{n+1}{2}\right)thand\left(\frac{n+1}{2}+1\right)thterm$. 

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 $\left(\frac{n}{2}+1\right)th$ term if n is even.

Concept:

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

${\mathrm{T}}_{\left(\mathrm{r}+1\right)}={}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}\times {\mathrm{x}}^{\mathrm{n}-\mathrm{r}}\times {\mathrm{y}}^{\mathrm{r}}$

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. $\left(\frac{\mathrm{n}}{2}+1\right){}^{\mathrm{t}\mathrm{h}}$ 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. ${\left(\frac{\mathrm{n}+1}{2}\right)}^{\mathrm{t}\mathrm{h}}$and ${\left(\frac{\mathrm{n}+3}{2}\right)}^{\mathrm{t}\mathrm{h}}$ are two middle terms.

Calculation:

Here, we have to find the middle terms in the expansion of ${(2\mathrm{x}+\frac{1}{\mathrm{x}})}^5$

Here n = 5 (n is odd number)

∴ Middle term =  ${\left(\frac{\mathrm{n}+1}{2}\right)}^{\mathrm{t}\mathrm{h}}$and ${\left(\frac{\mathrm{n}+3}{2}\right)}^{\mathrm{t}\mathrm{h}}$ = 3^rd and 4^th

T3 = T (2 + 1) = 5C2 × (2x) (5 - 2) × ${\left(\frac{1}{\mathrm{x}}\right)}^2$  and T4 = T (3 + 1) = 5C3 × (2x) (5 - 3) × ${\left(\frac{1}{\mathrm{x}}\right)}^3$ 

T3 =  5C2 × (2³x) and T4 = 5C3 × 2² × $\frac{1}{\mathrm{x}}$

T3 = 80x and T4 = $\frac{40}{\mathrm{x}}$

Hence the middle term of expansion is 80x and $\frac{40}{\mathrm{x}}$

Concept:

Expansion of (1 + x)n:

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

Calculation:

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

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

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

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

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

⇒ 4m² - 4m + 1 = 0

⇒ (2m - 1)² = 0

⇒ 2m - 1 = 0

∴ m = $\frac{1}{2}$

Concept:

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

• ${\mathrm{T}}_{\left(\mathrm{r}+1\right)}={}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}\times {\mathrm{x}}^{\mathrm{n}-\mathrm{r}}\times {\mathrm{y}}^{\mathrm{r}}$

Calculation:

Given expansion is ${\left(\sqrt{\mathrm{x}}+\frac{1}{3{\mathrm{x}}^2}\right)}^{10}$

General term = ${\mathrm{T}}_{\left(\mathrm{r}+1\right)}={}^{10}{\mathrm{C}}_{\mathrm{r}}\times {\mathrm{x}}^{\frac{10-\mathrm{r}}{2}}\times {\left(\frac{1}{3{\mathrm{x}}^2}\right)}^{\mathrm{r}}={}^{10}{\mathrm{C}}_{\mathrm{r}}\times 3^{-\mathrm{r}}\times {\mathrm{x}}^{\frac{10-5\mathrm{r}}{2}}$ 

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

i.e $\frac{10-5\mathrm{r}}{2}=0$

⇒ r = 2

Concept:

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

$T_{\left(r+1\right)}={}^n{C}_r\times x^{n-r}\times y^r$

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. $\left(\frac{n}{2}+1\right){}^{th}$ term is the middle term.

$T_{\left(\frac{n}{2}+1\right)}={}^n{C}_{\frac{n}{2}}\times x^{\frac{n}{2}}\times y^{\frac{n}{2}}$

• 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. ${\left(\frac{n+1}{2}\right)}^{th}$and ${\left(\frac{n+3}{2}\right)}^{th}$ 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 = $\left(\frac{n}{2}+1\right)=\left(\frac{4}{2}+1\right)=3rdterm$

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

T₃ = 6 × 4 × 9x² = 216 x²

Concept:

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

${\mathrm{T}}_{\left(\mathrm{r}+1\right)}={}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}\times {\mathrm{x}}^{\mathrm{n}-\mathrm{r}}\times {\mathrm{y}}^{\mathrm{r}}$

Expansion of (1 + x)ⁿ:

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

Calculation:

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

$$\begin{gathered} (4-5{\mathrm{x}}^2{)}^{-1/2}=4^{-1/2}{(1-\frac{5}{4}{\mathrm{x}}^2)}^{-1/2} \\ \text{As we know}(1+\mathrm{x}{)}^{\mathrm{n}}=1+\mathrm{n}\mathrm{x}+\frac{\mathrm{n}(\mathrm{n}-1)}{2!}{\mathrm{x}}^2+\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{3!}{\mathrm{x}}^3+.... \\ \therefore 4^{-1/2}{(1-\frac{5}{4}{\mathrm{x}}^2)}^{-1/2}=2^{-1}[1+(-\frac{5}{4}{\mathrm{x}}^2)\times (\frac{-1}{2})+...] \end{gathered}$$

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

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