Question types

Binomial Theorem question types

216 questions across 7 question groups — pick any mix to generate a MATHS paper with step-by-step answer keys.

216
Questions
7
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 term independent of x in the expansion of $\Big(9\text{x}-\frac{1}{3\sqrt{{x}}2}\Big)18,$ x > 0 , is ‘a’ times the corresponding binomial coefficient. Then ‘a’ is:
  • A
    $3$
  • B
    $\frac{1}{3}$
  • C
    $-\frac{1}{3}$
  • D
    $\text{None of these}$
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In the expansion of $\Big(\sqrt[3]4+\frac{1}{\sqrt[4]{6}}\Big)^{20},$
  • A
    The number of rational terms = 4
  • B
    The number of irrational terms = 19
  • C
    The middle term is irrational
  • D
    The number of irrational terms = 17
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State which of the statement in True or False.
The sum of the series $\sum\limits_{\text{r}=0}^{10}\ ^{20}\text{C}_\text{r}\ \text{is}\ 2^{19}+\frac{^{20}\text{C}_{10}}{2}$
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State which of the statement in True or False.
If the expansion of $\Big(\text{x}-\frac{1}{\text{x}^2}\Big)^{2\text{n}}$ contains a term independent of x, then n is a multiple of 2.
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Fill in the blank.
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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Q 263 Marks Question3 Marks
Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of ${\left( {\sqrt[4] 2 + \frac{1}{{\sqrt [4]{3} }}} \right)^n}$ is $\sqrt 6 :1$.
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Q 293 Marks Question3 Marks
If a and b are distinct integers, prove that a - b is a factor of an - bn, whenever n is a positive integer.
[Hint write an = (a – b + b)n and expand]
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