JEE Main 2026 · 5 Apr · Shift 1 · Mathematics Q23

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MathematicsInteger+4 / -1Question 23 of 75

If the sum of the coefficients of $x^7$ and $x^{14}$ in the expansion of $\left(\dfrac1{x^3}-x^4\right)^n$ , $x\ne0$ , is zero, then the value of $n$ is ________.

Your answer

Answer and worked solution

Official answer: 21

General term. $\binom nr\left(x^{-3}\right)^{n-r}(-x^4)^r=(-1)^r\binom nr x^{7r-3n}$ .

The two powers. $x^7$ needs $7r=3n+7$ and $x^{14}$ needs $7r'=3n+14$ , so $r'=r+1$ (and $7\mid n$ ).

Condition. $(-1)^r\left[\binom nr-\binom n{r+1}\right]=0$ gives $2r+1=n$ . With $r=\tfrac{3n}7+1$ : $\tfrac{6n}7+3=n$ , so $n=21$ .

Answer: 21

Topics:General and middle terms Properties of binomial coefficients

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JEE Main 2026 · 5 Apr · Shift 1, Mathematics question 23