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:

• 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.

Concept:

• Binomial Coefficient: In the expansion of \((1+x)^{10}\), the coefficient of \(x^{10-r}\) is given by \(a_r = \binom{10}{r}\).
• We are required to evaluate the expression: \(\sum_{r=1}^{10} r^3 \left( \frac{a_r}{a_{r-1}} \right)^2\)
• Using property of binomial coefficients: \(\frac{a_r}{a_{r-1}} = \frac{\binom{10}{r}}{\binom{10}{r-1}} = \frac{10 - r + 1}{r} = \frac{11 - r}{r}\)
• So, \(\left( \frac{a_r}{a_{r-1}} \right)^2 = \left( \frac{11 - r}{r} \right)^2\)
• Final expression becomes: \(\sum_{r=1}^{10} r^3 \cdot \left( \frac{11 - r}{r} \right)^2 = \sum_{r=1}^{10} (11 - r)^2\cdot r\)

Calculation:

We simplify:

\(\sum_{r=1}^{10} r \cdot (11 - r)^2\)

⇒ For r=1: \(1 \cdot 10^2 = 100 \) 

⇒ For r=2: \(2 \cdot 9^2 = 162 \) 

⇒ For r=3: \(3 \cdot 8^2 = 192 \)

⇒ For r=4: \( 4 \cdot 7^2 = 196 \)

⇒ For r=5: \(5 \cdot 6^2 = 180 \) 

⇒ For r=6: \(6 \cdot 5^2 = 150 \) 

⇒ For r=7: \(7 \cdot 4^2 = 112 \) 

⇒ For r=8: \(8 \cdot 3^2 = 72 \) 

⇒ For r=9:  \(9 \cdot 2^2 = 36 \)

⇒ For r=10: \(10 \cdot 1^2 = 10 \) 

⇒ Sum = 100 + 162 + 192 + 196 + 180 + 150 + 112 + 72 + 36 + 10 = 1210

∴ The value of the given summation is 1210.

Calculation:

Given,

We are given the equation:

\( \left( \frac{x+1}{x^{2/3} + 1 - x^{1/3}} - \frac{x+1}{x - x^{1/2}} \right)^{10}, x > 1 \)

Simplify the terms inside the bracket:

\( = \left( \frac{x+1}{(x^{1/3})^2 - x^{1/3} + 1} - \frac{x+1}{\sqrt{x}(\sqrt{x} - 1)} \right)^{10} \)

\( = \left( (x^{1/3} + 1) - \frac{\sqrt{x} + 1}{\sqrt{x}} \right)^{10} \)

\( = \left( x^{1/3} + 1 - 1 - x^{-1/2} \right)^{10} = \left( x^{1/3} - x^{-1/2} \right)^{10} \)

The general term T{r+1}\) is given by:

\( T_{r+1} = {}^{10}C_r (x^{1/3})^{10-r} (-x^{-1/2})^r = {}^{10}C_r (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}} \)

For the term independent of x, the exponent of x must be zero:

\( \frac{10-r}{3} - \frac{r}{2} = 0 \Rightarrow 2(10-r) - 3r = 0 \Rightarrow 20 - 5r = 0 \Rightarrow r = 4 \)

The required term is T₅:

\( T_5 = {}^{10}C_4 (-1)^4 x^0 = {}^{10}C_4 = \frac{10!}{4!6!} = 210 \)

∴ The term independent of x is 210.

Concept:

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

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

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

• If n is even, then 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. \(\rm {\left( {\frac{{n\; + \;1}}{2}} \right)^{th}}\;\)and \(\rm {\left( {\frac{{n\; + \;3}}{2}} \right)^{th}}\) are two middle terms.

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

Here n = 8 (n is even number)

∴ Middle term = \(\rm \left( {\frac{n}{2} + 1} \right) = \left( {\frac{8}{2} + 1} \right) =5th\;term\)

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

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

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:\(\left( {\frac{{n + 1}}{2}} \right)th\;and\;\left( {\frac{{n + 1}}{2} + 1} \right)th\;term\). 

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

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

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

• If n is even, then 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. \(\rm {\left( {\frac{{n\; + \;1}}{2}} \right)^{th}}\;\)and \(\rm {\left( {\frac{{n\; + \;3}}{2}} \right)^{th}}\) are two middle terms.

Calculation:

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

Here n = 5 (n is odd number)

∴ Middle term =  \(\rm {\left( {\frac{{n\; + \;1}}{2}} \right)^{th}}\;\)and \(\rm {\left( {\frac{{n\; + \;3}}{2}} \right)^{th}}\) = 3^rd and 4^th

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

T3 =  5C2 × (2³x) and T4 = 5C3 × 2² × \(\rm \frac 1 x\)

T3 = 80x and T4 = \(\rm \frac {40}{x}\)

Hence the middle term of expansion is 80x and \(\rm \frac {40}{x}\)

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

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

Calculation:

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

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

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

i.e \(\frac{{10{\rm{\;}} - 5{\rm{\;r}}}}{2} = 0\)

⇒ 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. \(\left( {\frac{n}{2} + 1} \right){{\rm{\;}}^{th}}\) 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 = \(\left( {\frac{n}{2} + 1} \right) = \left( {\frac{4}{2} + 1} \right) = 3rd\;term\)

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

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

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 = \(2^{-1} \times \frac{-5}{4} \times \frac{-1}{2} = \frac{5}{16}\)

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