Binomial Theorem Previous Year Questions (JEE Main 2026)

Use $\;{}^{36}C_{r+1} = \dfrac{36}{r+1}\,{}^{35}C_r.$ Substituting:

$$\frac{36}{r+1}\,{}^{35}C_r = \frac{6\,{}^{35}C_r}{k^2 - 3}.$$

Cancelling ${}^{35}C_r$ (nonzero for $0 \le r \le 35$):

$$\frac{36}{r+1} = \frac{6}{k^2 - 3} \;\Rightarrow\; k^2 - 3 = \frac{r+1}{6}.$$

So $k^2 = \dfrac{r+1}{6} + 3$ must be a perfect square, which needs $6 \mid (r+1)$.

• $r = 5:\; k^2 = \tfrac{6}{6} + 3 = 4 \Rightarrow k = \pm 2$
• $r = 35:\; k^2 = \tfrac{36}{6} + 3 = 9 \Rightarrow k = \pm 3$

Other multiples of $6$ give non-squares. Since $k \in \mathbb{Z}$, each valid $r$ yields two values of $k$, giving pairs $(5, \pm 2)$ and $(35, \pm 3)$ — a total of $4$ elements.

Answer: B (4)

The coefficients $1, 3, 3, 1$ are the expansion of $(1+x)^3$. Recall that $(1+x)^{30}(1+x)^3 = (1+x)^{33}$, and comparing coefficients of a fixed power gives a Vandermonde-type convolution:

$$\sum_{j=0}^{3}\binom{3}{j}\binom{30}{(33-r)-j} = \binom{33}{33-r}.$$

Writing the terms out, this is exactly

$${}^{30}C_{33-r} + 3\,{}^{30}C_{32-r} + 3\,{}^{30}C_{31-r} + {}^{30}C_{30-r} = {}^{33}C_{33-r} = {}^{33}C_r.$$

Hence $m = 33$.

Answer: C (33)

Coefficient of $x^3$ in $\displaystyle\sum_{j=3}^{99}(1+x)^j$ is $\displaystyle\sum_{j=3}^{99}\binom{j}{3} = \binom{100}{4}$ (hockey-stick identity).

Coefficient of $x^3$ in $(1+kx)^{100}$ is $\dbinom{100}{3}k^3$.

Total:

$$\binom{100}{4} + k^3\binom{100}{3} = \binom{100}{3}\left(\frac{97}{4} + k^3\right),$$

using $\dbinom{100}{4} = \dbinom{100}{3}\cdot\dfrac{97}{4}$.

Setting this equal to $\left(43n + \dfrac{101}{4}\right)\dbinom{100}{3}$:

$$\frac{97}{4} + k^3 = 43n + \frac{101}{4} \;\Rightarrow\; k^3 = 43n + 1.$$

We need the smallest $k \in \mathbb{N}$ with $k^3 \equiv 1 \pmod{43}$ and $n \in \mathbb{N}$ (so $n \ge 1$). Testing $k = 2,3,4,5$ gives $k^3 - 1 = 7, 26, 63, 124$, none divisible by $43$. For $k = 6$: $6^3 - 1 = 215 = 43\times 5$, so $n = 5$.

Thus $p = 6$, $n = 5$, and $p + n = 11$.

Answer: B (11)

General term:

$$T_{r+1} = \binom{18}{r}(9x)^{18-r}\left(-\frac{1}{3}x^{-1/2}\right)^{r}.$$

Power of $x$: $(18 - r) - \dfrac{r}{2} = 18 - \dfrac{3r}{2}$. For the independent term set this to $0$:

$$18 = \frac{3r}{2} \;\Rightarrow\; r = 12.$$

The coefficient is

$$\binom{18}{12}\,9^{6}\left(-\frac{1}{3}\right)^{12} = \binom{18}{12}\cdot\frac{9^{6}}{3^{12}} = \binom{18}{12}\cdot\frac{3^{12}}{3^{12}} = \binom{18}{12} = 18564.$$

Since $18564 = 221 \times 84$, we get $k = 84$.

Answer: A (84)

General term:

$$T_{r+1} = \binom{n}{r}\left(x^{-3}\right)^{n-r}\left(-x^{4}\right)^{r} = (-1)^{r}\binom{n}{r}\,x^{-3n + 7r}.$$

Coefficient of $x^7$: $-3n + 7r = 7 \Rightarrow r_1 = \dfrac{7 + 3n}{7}.$

Coefficient of $x^{14}$: $-3n + 7r = 14 \Rightarrow r_2 = \dfrac{14 + 3n}{7}.$

Both require $7 \mid 3n$, i.e. $7 \mid n$. Take $n = 7t$; then $r_2 = r_1 + 1$, so the coefficients are $(-1)^{r_1}\binom{n}{r_1}$ and $(-1)^{r_1+1}\binom{n}{r_1+1}$. Their sum is zero when

$$\binom{n}{r_1} = \binom{n}{r_1 + 1} \;\Rightarrow\; r_1 + (r_1 + 1) = n \;\Rightarrow\; n = 2r_1 + 1.$$

With $r_1 = \dfrac{7 + 3n}{7}$:

$$n = 2\cdot\frac{7 + 3n}{7} + 1 \;\Rightarrow\; 7n = 14 + 6n + 7 \;\Rightarrow\; n = 21.$$

Check: $r_1 = 10,\ r_2 = 11$, and $\binom{21}{10} = \binom{21}{11}$ with opposite signs cancel.

Answer: 21

General term:

$$T_{r+1} = \binom{10}{r}(2x^2)^{10-r}\left(x^{-1}\right)^{r} = \binom{10}{r}\,2^{10-r}\,x^{20 - 3r}.$$

For $x^2$: $20 - 3r = 2 \Rightarrow r = 6.$

Coefficient:

$$\binom{10}{6}\,2^{4} = 210 \times 16 = 3360.$$

Answer: B (3360)

$(1 + \alpha x)^{26}$ has $27$ terms, so the single middle term is the $14$th ($r = 13$):

$$\text{coeff} = \binom{26}{13}\alpha^{13}.$$

$(1 - \alpha x)^{28}$ has $29$ terms, so the middle term is the $15$th ($r = 14$):

$$\text{coeff} = \binom{28}{14}(-\alpha)^{14} = \binom{28}{14}\alpha^{14}.$$

Equating:

$$\binom{26}{13}\alpha^{13} = \binom{28}{14}\alpha^{14} \;\Rightarrow\; \alpha = \frac{\binom{26}{13}}{\binom{28}{14}}.$$

Now $\dbinom{28}{14} = \dbinom{26}{13}\cdot\dfrac{27\cdot 28}{14\cdot 14} = \dbinom{26}{13}\cdot\dfrac{756}{196} = \dbinom{26}{13}\cdot\dfrac{27}{7}.$

Therefore

$$\alpha = \frac{1}{27/7} = \frac{7}{27}.$$

Answer: D ($\tfrac{7}{27}$)

Factor the left side: $(1 - x^3)^{10} = (1-x)^{10}(1 + x + x^2)^{10}.$

Divide both sides by $(1-x)^{30}$ and let $y = \dfrac{x}{(1-x)^2}$:

$$\frac{(1+x+x^2)^{10}}{(1-x)^{20}} = \sum_{r=0}^{10} a_r\, y^{r}.$$

Note $1 + x + x^2 = (1-x)^2 + 3x$, so

$$\frac{(1+x+x^2)^{10}}{(1-x)^{20}} = \left(1 + \frac{3x}{(1-x)^2}\right)^{10} = (1 + 3y)^{10}.$$

Therefore $a_r = \dbinom{10}{r}3^{r}$. In particular:

$$a_9 = \binom{10}{9}3^{9} = 10\cdot 3^{9}, \qquad a_{10} = \binom{10}{10}3^{10} = 3^{10}.$$$$\frac{9a_9}{a_{10}} = \frac{9\cdot 10\cdot 3^{9}}{3^{10}} = \frac{90}{3} = 30.$$

Answer: 30

Consider the integral

$$\int_0^2 \frac{(1+x)^{12} - (1-x)^{12}}{2}\,dx.$$

The integrand keeps only odd powers of $x$: $\displaystyle \sum_{m=0}^{5}\binom{12}{2m+1}x^{2m+1}.$ Integrating from $0$ to $2$:

$$\int_0^2 \frac{(1+x)^{12}-(1-x)^{12}}{2}\,dx = \frac{1}{2}\left[\frac{(1+x)^{13} + (1-x)^{13}}{13}\right]_0^2 = \frac{3^{13} + (-1)^{13} - 2}{26} = \frac{3^{13} - 3}{26}.$$

Alternatively, integrating the even-part expansion $(1+x)^{12}+(1-x)^{12} = 2\sum_{m=0}^{6}\binom{12}{2m}x^{2m}$ gives, after evaluating the odd antiderivative at $x=2$:

$$\int_0^2\frac{(1+x)^{12}+(1-x)^{12}}{2}\,dx = \sum_{m=0}^{6}\binom{12}{2m}\frac{2^{2m+1}}{2m+1} = \frac{3^{13} + (-1)\cdot(-3^{13})}{26}\ldots$$

Direct evaluation of the given sum (terms $m = 1$ to $6$, i.e. excluding the $\binom{12}{0}$ term $\tfrac{2}{1}=2$):

$$S = \sum_{m=1}^{6}\frac{2^{2m+1}}{2m+1}\binom{12}{2m} = \frac{1594272}{26}.$$

Hence

$$26\,S = 1594272, \qquad 3^{13} = 1594323 \;\Rightarrow\; \alpha = 3^{13} - 26S = 1594323 - 1594272 = 51.$$

Answer: C (51)