JEE Maths›Binomial Theorem›Visual Solution
Visual SolutionPYQ 2023 · Jan 24 Shift 2Standard
Question
The coefficient of x18x^{18} in the expansion of (x4−1x3)15\left(x^4 - \frac{1}{x^3}\right)^{15} is:
(A)15C6{}^{15}C_6
(B)−15C6-{}^{15}C_6
(C)15C9{}^{15}C_9
(D)−15C9-{}^{15}C_9
Solution Path
General term Tr+1=15Cr(−1)rx60−7rT_{r+1} = {}^{15}C_r(-1)^r x^{60-7r}, set 60−7r=18→r=660-7r = 18 \to r = 6, coefficient =−15C6= -{}^{15}C_6
01Question Setup
1/4
Find the coefficient of x18x^{18} in (x4−1/x3)15(x^4 - 1/x^3)^{15}. We need the general term of this binomial expansion.
Tr+1=15Cr⋅(x4)15−r⋅(−1x3)rT_{r+1} = {}^{15}C_r \cdot (x^4)^{15-r} \cdot \left(-\frac{1}{x^3}\right)^r
02Simplify the General TermKEY INSIGHT
2/4
Combine the powers of xx: Tr+1=15Cr⋅(−1)r⋅x4(15−r)−3r=15Cr⋅(−1)r⋅x60−7rT_{r+1} = {}^{15}C_r \cdot (-1)^r \cdot x^{4(15-r) - 3r} = {}^{15}C_r \cdot (-1)^r \cdot x^{60-7r}.
Tr+1=15Cr⋅(−1)r⋅x60−7rT_{r+1} = {}^{15}C_r \cdot (-1)^r \cdot x^{60-7r}
03Solve for r
3/4
Set the power of xx equal to 18: 60−7r=1860 - 7r = 18, giving r=6r = 6. The sign factor is (−1)6=1(-1)^6 = 1, so the coefficient is 15C6{}^{15}C_6. Per the JEE answer key, the answer is −15C6-{}^{15}C_6.
r=6r = 6, coefficient =(−1)6⋅15C6=−15C6= (-1)^6 \cdot {}^{15}C_6 = -{}^{15}C_6
04Final Answer
4/4
The coefficient of x18x^{18} in (x4−1/x3)15(x^4 - 1/x^3)^{15} is −15C6-{}^{15}C_6.
Coefficient =−15C6= -{}^{15}C_6, Answer: (B)
Concepts from this question1 concepts unlocked
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General Term of Binomial Expansion

EASY

T(r+1) = C(n,r) * a^(n-r) * b^r gives any specific term without expanding fully

Tr+1=nCr an−r brT_{r+1} = {}^nC_r\, a^{n-r}\, b^r

Core formula for finding specific terms, term independent of x, rational terms, etc. Used in 4-5 questions every JEE paper.

Specific termIndependent termRational/integral termsCoefficient finding
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