Binomial TheoremeasyPYQ · JEE Main · 5 Apr 2026 · Shift 1 (Morning)Free

Coefficients of x⁷ and x¹⁴ Sum Zero: n = 21 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
If the sum of the coefficients of x7x^7 and x14x^{14} in the expansion of (1x3−x4)n\left(\dfrac{1}{x^3}-x^4\right)^n, x≠0x\ne0, is zero, then the value of nn is
Solution
Answer: 21 (± 0.01)
Step 1: Tr+1=nCr(1x3)n−r(−x4)r=nCr(−1)rx−3(n−r)+4r=nCr(−1)rx7r−3nT_{r+1}={}^nC_r\left(\dfrac{1}{x^3}\right)^{n-r}(-x^4)^r={}^nC_r(-1)^r x^{-3(n-r)+4r}={}^nC_r(-1)^r x^{7r-3n}. Step 2: 7r−3n=7⇒r1=7+3n77r-3n=7\Rightarrow r_1=\dfrac{7+3n}{7}; 7r−3n=14⇒r2=14+3n7=r1+17r-3n=14\Rightarrow r_2=\dfrac{14+3n}{7}=r_1+1. Consecutive terms. Step 3: Sum of coefficients zero:
nCr1(−1)r1+nCr1+1(−1)r1+1=0⇒(−1)r1(nCr1−nCr1+1)=0.{}^nC_{r_1}(-1)^{r_1}+{}^nC_{r_1+1}(-1)^{r_1+1}=0\Rightarrow(-1)^{r_1}\big({}^nC_{r_1}-{}^nC_{r_1+1}\big)=0.
⇒nCr1=nCr1+1⇒r1+(r1+1)=n.\Rightarrow{}^nC_{r_1}={}^nC_{r_1+1}\Rightarrow r_1+(r_1+1)=n.
 Step 4: 2r1+1=n2r_1+1=n, r1=7+3n7r_1=\dfrac{7+3n}{7}: 2⋅7+3n7+1=n⇒2(7+3n)+7=7n⇒21+6n=7n⇒n=212\cdot\dfrac{7+3n}{7}+1=n\Rightarrow 2(7+3n)+7=7n\Rightarrow 21+6n=7n\Rightarrow n=21. Correct answer: 21
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