Question types

Binomial Theorem question types

41 questions across 6 question groups — pick any mix to generate a Maths paper with step-by-step answer keys.

41
Questions
6
Question groups
5
Question types
Sample Questions

Binomial Theorem questions

One sample from each question group in this chapter. Select any group above to see the full set with answer keys.

The two successive terms in the expansion of $(1 + x)^{24}$ whose coefficients are in the ratio $1 : 4$ are: $[$Hint:$\frac{^{24}\text{C}_\text{r}}{^{24}\text{C}_{\text{r}+1}}=\frac{1}{4}\ \frac{\text{r}+1}{24-\text{r}}\ \frac{1}{4}\Rightarrow4\text{r}+4=24-4\Rightarrow\text{r}=4]$
  • A
    $3^{rd} and 4^{th}.$
  • B
    $4^{th} and 5^{th}.$
  • $5^{th} and 6^{th}.$
  • D
    $6^{th} and 7^{th}.$

Answer: C.

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If the coefficients of $2^{nd}, 3^{rd}$ and the $4^{th}$ terms in the expansion of $(1 + x)^n$ are in $A.P.,$ then value of n is: $[$Hint: $2^nC_2 = ^nC_1 + ^nC_3 \Rightarrow n^2 - 9n + 14 = 0 \Rightarrow n = 2\  $or $7.]$
  • A
    $2.$
  • $7.$
  • C
    $11.$
  • D
    $14.$

Answer: B.

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Choose the correct answer.
The total number of terms in the expansion of $(x + a)^{100} + (x - a)^{100}$ after simplification is:
  • A
    $50.$
  • B
    $202.$
  • $51.$
  • D
    None of these.

Answer: C.

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The coefficient of $x^n$ in the expansion of $(1 + x)^{2n} and (1 + x)^{2n - 1}$ are in the ratio. $[$Hint: $^{2\text{n}}\text{C}_\text{n} : \ ^{2\text{n} - 1}\text{C}_\text{n}]$
  • A
    $1 : 2.$
  • B
    $1 : 3.$
  • C
    $3 : 1.$
  • $2 : 1.$

Answer: D.

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If A and B are coefficient of $x^n$ in the expansions of $(1 + x)^{2n}$ and $(1 + x)^{2n – 1}$ respectively, then $\frac{\text{A}}{\text{B}}$ equals:
$[$Hint: $\frac{\text{A}}{\text{B}}=\frac{^{2\text{n}}\text{C}_\text{n}}{^{2\text{n}-1}\text{C}_\text{n}}=2]$
  • A
    $1.$
  • $2.$
  • C
    $\frac{1}{2}.$
  • D
    $\frac{1}{\text{n}}.$

Answer: B.

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Q 133 Marks Question3 Marks
Show that the middle term in the expansion of $\Big(\text{x}-\frac{1}{\text{x}}\Big)^{2\text{n}}$ is $\frac{1\times3\times5\times....(2\text{n}-1)}{\text{n}!}\times(-2)^\text{n}.$
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Q 153 Marks Question3 Marks
Find n in the binomial $\Big(3\sqrt{2}+\frac{1}{3\sqrt{3}}\Big)^\text{n}$ if the ratio of $7^{th}$ term from the beginning to the $7^{th}$ term from the end is $\frac{1}{6}.$
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The coefficient of $a^{-6}b^4$ in the expansion of $\Big(\frac{1}{\text{a}}-\frac{2\text{b}}{3}\Big)^{10}$ is ___________.
$[$Hint: $\text{T}_5=\ ^{10}\text{C}_4\Big(\frac{1}{\text{a}}\Big)^\text{b}\Big(\frac{-2\text{b}}{3}\Big)^4=\frac{1120}{27}\text{a}^{-6}\text{b}^4]$
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If the seventh terms from the beginning and the end in the expansion of $\Big(3\sqrt{2}+\frac{1}{3\sqrt{3}}\Big)^\text{n}$ are equal, then n equals _____________.
[Hint: $\text{T}_7=\text{T}_{\text{n}-7+2}\Rightarrow\ ^\text{n}\text{C}_6\Big(2^\frac{1}{3}\Big)^{\text{n}-6}\bigg(\frac{1}{3^\frac{1}{3}}\bigg)^6$ $=\ ^\text{n}\text{C}_{\text{n}-6}\Big(2^\frac{1}{3}\Big)^6\bigg(\frac{1}{3^\frac{1}{3}}\bigg)^{\text{n}-6}$
$\Rightarrow\Big(2^\frac{1}{3}\Big)^{\text{n}-12}=\bigg(\frac{1}{3^{\frac{1}{3}}}\bigg)^{\text{n}-12}\Rightarrow$ only problem when $\text{n}-12=0\Rightarrow\text{n}=12]$
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Find the sixth term of the expansion $\Big(\text{y}^\frac{1}{2}+\text{x}^\frac{1}{3}\Big)^\text{n},$ if the binomial coefficient of the third term from the end is $45.$
$[$Hint: Binomial coefficient of third term from the end $=$ Binomial coefficient of third term from beginning $= ^nC_2.]$
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In the expansion of $(x + a)^n$ if the sum of odd terms is denoted by $O$ and the sum of even term by $E$.
Then prove that,
  1. $\text{O}^2 – \text{E}^2 = (\text{x}^2 – \text{a}^2 )^\text{n}$
  2. $4\text{OE} = (\text{x} + \text{a})^{2\text{n}} – (\text{x} – \text{a})^{2\text{n}}$
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