JEE Main 2026 · 6 Apr · Shift 2, Mathematics question 22

Rewrite. Divide by $(1-x)^{30}$ and put $u=\dfrac x{(1-x)^2}$:
$$\left(\frac{1+x+x^2}{(1-x)^2}\right)^{10}=\sum a_ru^r.$$

Key identity. $1+x+x^2=(1-x)^2+3x$, so the left side is $(1+3u)^{10}$ and $a_r=\binom{10}r3^r$.

Ratio.
$$\frac{9a_9}{a_{10}}=\frac{9\cdot10\cdot3^9}{3^{10}}=30.$$

Answer: 30