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Binomial Theorem — BITSAT Mathematics PYQs

34 previous year questions from Binomial Theorem with answers and solutions. Numbered list, year tags, and one-tap solutions — built for serious JEE / NEET practice.

34 questionsMathematicsSolutions on every page
1

The coefficient of x^2 term in the binomial expansion of ( 1 3 x^ 1 2 +x^ -1 4 )¹⁰ is

BITSAT 2024 Solution
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2

The coefficient of x^n in the expansion of e^ 7 x +e^x e^ 3 x is

BITSAT 2024 Solution
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3

The coefficient of the highest power of x in the expansion of (x+ x^2-1 )^8+ (x- x^2-1 )^8

BITSAT 2024 Solution
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4

If the 17^ th and the 18^ th terms in the expansion of (2+a)⁵⁰ are equal, then the coefficient of x³⁵ in the expansion of (a+x)⁻² is

BITSAT 2024 Solution
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5

If x¹⁸ occurs in the rth term in the expansion of (x^4+ 1 x^3 )¹⁵ , then what is the value of r ?

BITSAT 2023 Solution
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6

The middle term in the expansion of ( 10 x + x 10 )¹⁰ is

BITSAT 2023 Solution
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7

The number of terms in the expansion of (1+5 2 x)^9+(1-5 2 x)^9 , is

BITSAT 2022 Solution
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8

If the sum of the coefficients in the expansion of (x+y)^n is 1024 , then the value of the greatest coefficient in the expansion is

BITSAT 2022 Solution
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9

The coefficient of the term independent of x in the expansion of ( x 3 + 3 2 x² )¹⁰ is

BITSAT 2021 Solution
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10

Coefficient of x¹³ in the expansion of (1-x)⁵ (1+x+x²+x³ )⁴ is

BITSAT 2021 Solution
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11

If a, b, c are three natural numbers in AP and a + b + c =21 then the possible number of values of the ordered triplet ( a , b , c ) is

BITSAT 2020 Solution
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12

Let 2 x 2 + 3 x + 4 10 = ∑ r = 0 20 a r x r , then the value of a₈ a₁₂ , is

BITSAT 2019 Solution
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13

If 1 + x + x 2 20 = ∑ r = 0 40 a r · x r , then ∑ r = 0 39 - 1 r · a r · a r + 1 equal to

BITSAT 2019 Solution
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14

If the coefficient of x 3 and x 4 in the expansion of 1 + a x + b x 2 1 - 2 x 18 in powers of x are both zero, then a , b is equal to

BITSAT 2019 Solution
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15

In the expansion of 1 + x + x 3 + x 4 10 , the coefficient of x 4 is

BITSAT 2019 Solution
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16

The coefficient of x - n in 1 + x n 1 + 1 x n is

BITSAT 2018 Solution
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17

The greatest term in the expansion of 3 1 + 1 3 20 is

BITSAT 2018 Solution
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18

The coefficient of x 5 in the expansion of 1 + x 21 + 1 + x 22 + … + 1 + x 30 is

BITSAT 2017 Solution
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19

The number of values of satisfying the equation ³⁹ C _ 3 r -1 - ³⁹ C _ r ² = ³⁹ C _ r ²-1 - ³⁹ C _ 3 r is

BITSAT 2016 Solution
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20

If _ r=0 ^ n r+2 r+1 ^ n C_ r = 2⁸-1 6 , then n=

BITSAT 2016 Solution
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21

If the sum of odd numbered terms and the sum of even numbered terms in the expansion of (x+a)^ n are A and B respectively, then the value of (x²- . .a² )^ n is

BITSAT 2016 Solution
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22

If the third term in the expansion of [ x + x ^ ₁₀ x ]⁵ is 10⁶, then x may be

BITSAT 2016 Solution
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23

If e^ x +e^ 5 x e^ 3 x =a₀+a₁ x+a₂ x²+a₃ x³+ then the value of 2 a₁+2³ a₃+2⁵ a₅+ is

BITSAT 2015 Solution
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24

The coefficient of x ⁴ in the expansion of (1+x+x²+x³ )¹¹, is

BITSAT 2014 Solution
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25

If T ₀, ~T ₁, ~T ₂ T _ n represent the terms in the expansion of (x+a)^ n , then ( T ₀- T ₂+ T ₄- . )²+ ( T ₁- T ₃+ T ₅- . . )²=

BITSAT 2014 Solution
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26

The coefficient of the middle term in the expansion of (2+3 x )⁴ is:

BITSAT 2013 Solution
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27

If C ₀, C ₁, C ₂, . C _ n denote the binomial coefficientsin the expansion of (1+ x )^ n , then the value of C ₀+ ( C ₀+ C ₁ )+ ( C ₀+ C ₁+ C ₂ )+ + (C₀+C₁+ . .+C_ n-1 )

BITSAT 2013 Solution
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28

The coefficient of x²⁰ in the expansion of (1+x² )⁴⁰ (x²+2+ 1 x² )⁻⁵ is

BITSAT 2012 Solution
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29

For (n N , x^ n+1 +(x+1)^ 2 n-1 ) is divisible by

BITSAT 2011 Solution
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30

The term independent of (x ) in the expansion of ( (9 x - 1 3 x )¹⁸, x > 0 ), is a times the corresponding binomial coefficient. Then a is

BITSAT 2011 Solution
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31

In the binomial ( (2^ 1 / 3 +3^ -1 / 3 )^ n ), if the ratio of the seventh term from the beginning of the expansion to the seventh term from its end is (1 / 6 ), then (n ) equal to

BITSAT 2011 Solution
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32

With 17 consonants and 5 vowels the number of words of four letters that can be formed having two different vowels in the middle and one consonant, repeated or different at each en

BITSAT 2010 Solution
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33

The general solution of (8 ^2 x 2 =1+ x ) is

BITSAT 2009 Solution
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34

The letters of the word TOUGH are written in all possible orders and these words are written out as in a dictionary, then the rank of the word TOUGH is

BITSAT 2009 Solution
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