JEE Main 2026 — Binomial Theorem Question with Solution

The given equation is:

$(1-x^3{)}^{10}=\sum _{r=0}^{10}{a}_r{x}^r(1-x{)}^{30-2r}$

Using the algebraic identity $1-x^3=(1-x)(1+x+x^2)$, the left hand side can be written as:

$(1-x{)}^{10}(1+x+x^2{)}^{10}=\sum _{r=0}^{10}{a}_r{x}^r(1-x{)}^{30-2r}$

Dividing both sides by $(1-x{)}^{30}$:

$\frac{(1-x{)}^{10}(1+x+x^2{)}^{10}}{(1-x{)}^{30}}=\sum _{r=0}^{10}{a}_r\frac{x^r}{(1-x{)}^{2r}}$

${\left(\frac{1+x+x^2}{(1-x{)}^2}\right)}^{10}=\sum _{r=0}^{10}{a}_r{\left(\frac{x}{(1-x{)}^2}\right)}^r$

Let $y=\frac{x}{(1-x{)}^2}$.

The term inside the bracket on the left hand side can be simplified as:

$\frac{1+x+x^2}{(1-x{)}^2}=\frac{1-2x+x^2+3x}{(1-x{)}^2}=\frac{(1-x{)}^2+3x}{(1-x{)}^2}=1+3\left(\frac{x}{(1-x{)}^2}\right)=1+3y$

Substituting this into the equation, we get:

$(1+3y{)}^{10}=\sum _{r=0}^{10}{a}_r{y}^r$

Using the binomial expansion, $(1+3y{)}^{10}=\sum _{r=0}^{10}{}^{10}{C}_r(3y{)}^r=\sum _{r=0}^{10}{}^{10}{C}_r{3}^r{y}^r$.

Comparing the coefficients of $y^r$ on both sides, we obtain:

$a_r={}^{10}{C}_r{3}^r$

For $r=9$ and $r=10$:

$a_9={}^{10}{C}_9{3}^9=10\times 3^9$

$a_{10}={}^{10}{C}_{10}{3}^{10}=1\times 3^{10}=3^{10}$

Therefore, the required value is:

$\frac{9a_9}{a_{10}}=\frac{9\times 10\times 3^9}{3^{10}}=\frac{90\times 3^9}{3\times 3^9}=\frac{90}{3}=30$

Answer: $30$