Special Terms of Binomial Expansion MCQ Quiz - Objective Question with Answer for Special Terms of Binomial Expansion - 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.

7Concept:

Binomial Expansion:

• The binomial theorem is used to expand expressions of the form \((1 + x)^n\).
• The general term in the expansion of \((1 + x)^n\) is given by \(T_k = C(n, k) \cdot x^k\), where ⁿC_k is the binomial coefficient.
• To find the coefficient of x^3, we identify the corresponding terms from the expansion and set the coefficient equal to 35.

Calculation:

Given the expansion of \((1 + x)^p \cdot (1 + x)^q\), we have:

The coefficient of x³ in  the expansion is 35.

We use the binomial expansion formula for the term x³

⇒ (p+ q)_c3 = 35 = ⁷C₃

⇒ p+q =7

∴ the correct answer is Option C

Concept:

• The general term in a binomial expansion of (a + b)n is given by: \(T_{r+1}={}^nC_ra^{n-r}b^r\)
• If sin θ = sin α ⇒ θ = \(\rm n\pi +(-1)^n \alpha\)

Calculation:

Given, middle term of \(\left(\frac{1}{x}+x \sin x\right)^{10}\)is equal to \(7\frac{7}{8}\).

Since, n = 10

⇒ Middle term = \(\left(\frac{n}{2}+1\right)^{th}\) term = 6^th term.

∴ T₆ = T₅₊₁

= \({}^{10}C_5\left(\frac{1}{x} \right )^{10-5}(x\sin x)^5\)

⇒ \(\frac{63}{8}\) = \({}^{10}C_5(\sin x)^5\)

⇒ \(\frac{63}{8}\) = \(\frac{10!}{5!5!}\) sin⁵x

⇒ \(\frac{63}{8}\) = 252 sin5x

⇒ sin5x = \(\frac{1}{32}\)

⇒ sin x = \(\frac{1}{2}\) = sin\(\frac{π}{6}\)

∴ x = \(\rm nπ +(-1)^n \frac{π}{6}\)

Calculation

Given:

A = Coefficient of 30th term in (1 + x)2n-1 = 2n–1C29

B = Coefficient of 12th term in (1 + x)2n-1 = 2n–1C11 

2A = 5B

⇒ \(2 \frac{(2 n-1)!}{29!(2 n-30)!}=5 \frac{(2 n-1)!}{(2 n-12)!11!}\)

⇒ \(\frac{1}{29 \ldots 12 \cdot 5}=\frac{1}{(2 n-12)(2 n-13) \ldots(2 n-29)^{2}}\)

⇒ \(\frac{1}{30 \cdot 29 \ldots 12}=\frac{1}{(2 n-12)(2 n-13) \ldots(2 n-29) 12}\)

⇒ 2n – 12 = 30

⇒ n = 21

Hence option 2 is correct

Concept:

General term: General term in the expansion of (a + b)n is given by

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

Calculation:

We know that Tr+1 = Cr an-r br.

In the given binomial expression \(\rm\left(x-\frac1x\right)^{10}\), n = 10, a = x and b = \(\rm\frac{-1}{x}\).

∴ Tr+1 = ¹⁰Cr x^10-r \(\rm\left(\frac{-1}{x}\right)^r\) = 10Cr (-1)^r x10-2r

For the term to be independent of x, we must have 10 - 2r = 0.

⇒ r = 5.

The required term is:

10C5 (-1)⁵ = - ¹⁰C₅.

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:

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:

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:

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

General term: General term in the expansion of (x + y) ⁿ is given by: \({{\rm{T}}_{\left( {{\rm{r\;}} + {\rm{\;}}1} \right)}} = {\rm{\;}}{{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}} \times {{\rm{x}}^{{\rm{n}} - {\rm{r}}}} \times {{\rm{y}}^{\rm{r}}}\)

Calculation:

We have to find term independent of x in \({\left( {{{\rm{x}}^2} - {\rm{\;}}\frac{1}{{{{\rm{x}}^3}}}} \right)^{10}}\)

We know that,

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

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

\( = {\rm{\;}}{\left( { - 1} \right)^{\rm{r}}} \times {{\rm{\;}}^{10}}{{\rm{C}}_{\rm{r}}} \times {\left( {\rm{x}} \right)^{20 - 2{\rm{r}}}} \times {\left( {{{\rm{x}}^3}} \right)^{ - {\rm{r}}}}\)

\( = {\rm{\;}}{\left( { - 1} \right)^{\rm{r}}} \times {{\rm{\;}}^{10}}{{\rm{C}}_{\rm{r}}} \times {\left( {\rm{x}} \right)^{20 - 2{\rm{r}}}} \times {\left( {\rm{x}} \right)^{ - 3{\rm{r}}}}\)

\( = {\rm{\;}}{\left( { - 1} \right)^{\rm{r}}} \times {{\rm{\;}}^{10}}{{\rm{C}}_{\rm{r}}} \times {\left( {\rm{x}} \right)^{20 - 5{\rm{r}}}}\)

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

Therefore, 20 – 5r = 0

⇒ r = 4

\({{\rm{T}}_{\left( {4{\rm{\;}} + {\rm{\;}}1} \right)}} = {\rm{\;\;}}{\left( { - 1} \right)^4} \times {{\rm{\;}}^{10}}{{\rm{C}}_4} = {\rm{\;}}{{\rm{\;}}^{10}}{{\rm{C}}_4}{\rm{\;}}\)

Concept:

General term: General term in the expansion of (a + b)n is given by

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

Calculation:

We know that Tr+1 = ⁿCr an-r br.

In the given binomial expression (3 + ax)9, n = 9, a = 3 and b = ax.

∴ Tr+1 = 9Cr 39-r (ax)r = 9Cr 3⁹ \(\rm\left(\frac{a}{3}\right) ^r\) xr.

For the coefficients of x2 and x³, we must have r = 2 and 3 respectively.

⇒ 9C2 39 \(\rm\left(\frac{a}{3}\right) ^2\) = 9C3 39 \(\rm\left(\frac{a}{3}\right) ^3\)

⇒ a = \(\frac97\).

Concept:

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

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

Calculation:

We have to find term independent of x in \(\rm \left( x^2 - \frac 1 x \right)^9\)

As we know, \({{\rm{T}}_{\left( {{\rm{r\;}} + {\rm{\;}}1} \right)}} = {\rm{\;}}{{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}} × {{\rm{x}}^{{\rm{n}} - {\rm{r}}}} × {{\rm{y}}^{\rm{r}}}\)

⇒  \({{\rm{T}}_{\left( {{\rm{r\;}} + {\rm{\;}}1} \right)}} = {\rm{\;}}{{\rm{\;}}^{\rm{9}}}{{\rm{C}}_{\rm{r}}} × ({{\rm{x^2}})^{{\rm{9}} - {\rm{r}}}} \rm × ({{\rm\frac{-1}{x}})^{\rm{r}}}\)

= ⁹C_r x^{18 - 2r} (-1)^r x^-r

= (-1)^r 9C_r x^{18 - 3r}

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

Therefore, 18 - 3r = 0

∴ r = 6

Hence the value is (-1)^6 9C₆ = 84

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.

(1−x)n=∑k=0n(nk)1n−k(−x)k(1−x)n=∑k=0n(nk)1n−k(−x)k(1−x)n=∑k=0n(nk)1n−k(−x)

Calculation:

Given:

(1 + 4x + 4x2)5

⇒ [(1 + 2x)²]⁵

⇒ (1+ 2x)¹⁰ 

Here n = 10 (n is even number)

∴ Middle term = \(\left( {\frac{n}{2} + 1} \right) = \left( {\frac{10}{2} + 1} \right) \) = 6th term 

Middle Term, T₆ =  T_{5 + 1} = ¹⁰C₅ (1)⁵ (2x)⁵

⇒ \(\frac {10!}{5!5!}\)× 32x⁵

⇒ 8064 x⁵

∴ The coefficient of the middle term in the expansion of (1 + 4x + 4x2)5  is 8064.

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 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 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 \(\rm \left(x + \frac 1 x \right)^{10}\)

Here n = 10 (n is even number)

∴ Middle term = \(\rm \left( {\frac{n}{2} + 1} \right) = \left( {\frac{10}{2} + 1} \right) =6th\;term\)

T₆ = T (5 + 1) = ¹⁰C₅ × (x) (10 - 5) × \(\rm \left(\frac {1}{x}\right)^5\)

T₆ =  ¹⁰C5

Concept:

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

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

In the expansion of (x + y)n the number of terms is (n + 1)

From the end (n + 1)^th term is 1st term and n^th term is 2 term 

Calculation:

In the expansion of \(\rm (2x - \frac {x} {2})^n\) ,n^th term from the end of an expansion is 2nd term 

T_r+1 = ⁿC_r (2x)^{(n - r)} \(\rm (\frac {-x} {2})^r\)

T₂ =  ⁿC₁.(2x)^{(n - 1)} \(\rm (\frac {-x} {2})\)

= \(\rm - n \times 2^{(n - 1 - 1)} \times x^{(n - 1 +1)}\)

= \(\rm - n \times 2^{(n - 2)} \times x^{n}\)

= \(\rm - nx^n​​​. 2^{(n - 2)}\)