Binomial Theorem – JEE Main MathematicsPractice Questions & PYQs

The binomial expansion $(1+x)^n$ generates a series where the general term is defined by the following equation:
$T_{r+1} = \frac{n(n-1)(n-2)\cdot s(n-r+1)}{r!} x^r$
Because $x$ is positive, the sign of each term depends entirely on the product of the factors in the numerator, as the denominator $r!$ and $x^r$ are always positive. With $n = \frac{27}{5} = 5.4$, the terms in the expansion remain positive until a factor $(n-r+1)$ becomes less than zero, which occurs when $5.4 - r + 1 < 0$. This inequality simplifies to $r > 6.4$, and since $r$ must be an integer, the smallest value that satisfies this condition is $r=7$. Consequently, the eighth term, $T_{7+1}$, is the first negative term in the expansion.

The expression $(1+0.0001)^{10000}$ can be rewritten as $(1+\frac{1}{10^4})^{10^4}$, allowing us to apply the binomial theorem to expand the terms as $1 + 10^4 \times \frac{1}{10^4} + \frac{10^4(10^4-1)}{2!} \times \frac{1}{(10^4)^2} + \dots$. This series simplifies to $1 + 1 + \frac{1}{2!}(1 - \frac{1}{10^4}) + \dots$, which is strictly less than the infinite series $1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots$ that defines Euler's number $e$. Because $e \approx 2.71828$ and the value of our expression is slightly less than this, the smallest positive integer greater than the result is 3.

The sum of coefficients in a binomial expansion is determined by setting each variable to 1, which simplifies $(a+b)^n$ to $2^n$. Given that the total sum is 4096, we set up the equation $2^n = 4096$. Since $4096 = 2^{12}$, it follows that $n = 12$.

For an even exponent such as 12, the greatest coefficient in the expansion corresponds to the middle term, defined by ${}^n C_{n/2}$. With 12 as our index, we must calculate the value of ${}^{12} C_6$. The evaluation is as follows:

$\frac{12!}{6! \times 6!} = 924$

This result is the largest coefficient in the binomial expansion.

The expansion of $(1+x)^n$ is governed by the Binomial Theorem, where the coefficient of the term $x^r$ is defined by the binomial coefficient ${}^{n}C_r$. A fundamental property of these coefficients is symmetry, given by the identity ${}^{n}C_r = {}^{n}C_{n-r}$, which implies that selecting $r$ items from a set of $n$ is equivalent to excluding $r$ items from that set.

When expanding $(1+x)^{p+q}$, the coefficients for $x^p$ and $x^q$ are expressed as ${}^{p+q}C_p$ and ${}^{p+q}C_q$, respectively. By applying the symmetry identity with $n = p+q$, the expression ${}^{p+q}C_q$ transforms into ${}^{p+q}C_{(p+q)-q}$, which simplifies directly to ${}^{p+q}C_p$. Because both expressions reduce to the same value, the coefficients of $x^p$ and $x^q$ are equal.

The core principle for equal binomial coefficients ${}^N C_a = {}^N C_b$ is that the sum of the lower indices must equal the total exponent, such that $a + b = N$, provided the terms are distinct. Given the expansion $(1+x)^{2n}$, the coefficient of the $(r+2)^{\text{th}}$ term is expressed as ${}^{2n}C_{r+1}$ and the coefficient of the $3r^{\text{th}}$ term is expressed as ${}^{2n}C_{3r-1}$.

Setting these two values equal, we have ${}^{2n}C_{r+1} = {}^{2n}C_{3r-1}$. Applying the sum property, the indices must satisfy $(r+1) + (3r-1) = 2n$. Simplifying the left side yields $4r = 2n$, which leaves the final relationship as $n = 2r$.

The middle term of a binomial expansion $(a+x)^n$ with an even exponent $n$ occurs at position $r+1$ where $r = n/2$. For the expansion $\left(\frac{1}{\sqrt{6}}+\beta x\right)^4$, the middle term is the third term, yielding the coefficient $\binom{4}{2} (\frac{1}{\sqrt{6}})^2 \beta^2 = 6 \cdot \frac{1}{6} \cdot \beta^2 = \beta^2$. In the expansion $(1-3\beta x)^2$, the middle term is the second term, resulting in the coefficient $\binom{2}{1} (-3\beta) = -6\beta$. For the final expansion $\left(1-\frac{\beta}{2}x\right)^6$, the middle term is the fourth term, giving the coefficient $\binom{6}{3} (-\frac{\beta}{2})^3 = 20 \cdot (-\frac{\beta^3}{8}) = -\frac{5\beta^3}{2}$.

Since these three coefficients form an arithmetic progression, the second term must be the arithmetic mean of the first and third terms, expressed as $2 \times (-6\beta) = \beta^2 - \frac{5\beta^3}{2}$. This simplifies to $-12\beta = \beta^2 - 2.5\beta^3$. Since $\beta > 0$, we can divide by $\beta$ to obtain $-12 = \beta - 2.5\beta^2$, which rearranges into the following quadratic equation:
$5\beta^2 - 2\beta - 24 = 0$
Factoring the equation yields $(5\beta - 12)(\beta + 2) = 0$, and given $\beta > 0$, we find $\beta = \frac{12}{5}$. The common difference $d$ of the arithmetic progression is defined as the difference between the second and first terms, giving $d = -6\beta - \beta^2$. Substituting this into the expression $50 - \frac{2d}{\beta^2}$ provides $50 - \frac{2(-6\beta - \beta^2)}{\beta^2}$, which simplifies to $50 + \frac{12\beta}{\beta^2} + \frac{2\beta^2}{\beta^2}$, or $52 + \frac{12}{\beta}$. Using $\beta = \frac{12}{5}$, we calculate $52 + \frac{12}{(12/5)} = 52 + 5$, which results in 57.

Expanding the given binomial expression relies on isolating the specific term that satisfies the power requirement for the variable $x$. For the expansion $\left(\frac{\sqrt{x}}{5^{\frac{1}{4}}} + \frac{\sqrt{5}}{x^{\frac{1}{3}}}\right)^{60}$, the general term is defined by $\binom{60}{r} \cdot \left(\frac{x^{1/2}}{5^{1/4}}\right)^{60-r} \cdot \left(\frac{5^{1/2}}{x^{1/3}}\right)^{r}$. Setting the total exponent of $x$ to $10$ requires solving the equation $\frac{60-r}{2} - \frac{r}{3} = 10$, which leads to $\frac{180-5r}{6} = 10$, simplifying to $5r = 120$, resulting in $r = 24$.

Once the value of $r$ is determined, the constant coefficient consists of the combination $\binom{60}{24}$ multiplied by the remaining factors of 5 from the expansion. These base factors are $5^{-\frac{60-24}{4}} \cdot 5^{\frac{24}{2}}$, which equals $5^{-9} \cdot 5^{12}$, or $5^3$. To determine the total exponent $k$, one must evaluate the exponent of 5 within the binomial coefficient $\binom{60}{24} = \frac{60!}{24! \cdot 36!}$ using Legendre's formula. The exponent of 5 in $60!$ is $\lfloor \frac{60}{5} \rfloor + \lfloor \frac{60}{25} \rfloor = 12 + 2 = 14$. Similarly, the exponent of 5 in $24!$ is $\lfloor \frac{24}{5} \rfloor = 4$, and in $36!$ it is $\lfloor \frac{36}{5} \rfloor + \lfloor \frac{36}{25} \rfloor = 7 + 1 = 8$.

Subtracting the combined exponents of the denominator factorials from the exponent of the numerator factorial gives $14 - (4 + 8) = 2$. Adding this result to the power of 3 already found in the constant terms of the expansion, the total value for $k$ is $2 + 3 = 5$.

The binomial expansion of $\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{n}$ consists of $n+1$ terms. The fifth term from the beginning is given by $T_5 = {}^n C_4 \left(\sqrt[4]{2}\right)^{n-4} \left(\frac{1}{\sqrt[4]{3}}\right)^4$, and the fifth term from the end corresponds to the $(n-3)$-th term from the beginning, expressed as $T_{n-3} = {}^n C_{n-4} \left(\sqrt[4]{2}\right)^4 \left(\frac{1}{\sqrt[4]{3}}\right)^{n-4}$. Because the binomial coefficients ${}^n C_4$ and ${}^n C_{n-4}$ are equal, their ratio simplifies to the ratio of the variable components, resulting in $\frac{2^{(n-4)/4} \cdot 3^{-1}}{2^1 \cdot 3^{-(n-4)/4}} = 6^{(n-8)/4}$. Equating this expression to the given ratio $\sqrt[4]{6}$, or $6^{1/4}$, yields the equation $\frac{n-8}{4} = \frac{1}{4}$, which confirms that $n=9$.

With $n=9$, the sixth term from the beginning is $T_6 = {}^9 C_5 \left(\sqrt[4]{2}\right)^{9-5} \left(\frac{1}{\sqrt[4]{3}}\right)^5$. Evaluating the combination ${}^9 C_5$ gives 126, so the term becomes $126 \cdot \left(\sqrt[4]{2}\right)^4 \cdot \left(\frac{1}{\sqrt[4]{3}}\right)^5$. Simplifying further, $126 \cdot 2 \cdot \frac{1}{3 \cdot \sqrt[4]{3}}$ reduces to $\frac{252}{3 \cdot \sqrt[4]{3}}$, which is equivalent to $\frac{84}{\sqrt[4]{3}}$. Comparing this result to the given form $\frac{\alpha}{\sqrt[4]{3}}$, it is clear that $\alpha = 84$.

The binomial coefficients of the second, third, and fourth terms in the expansion, $\binom{m}{1}$, $\binom{m}{2}$, and $\binom{m}{3}$, represent the first, third, and fifth terms of an arithmetic progression. This relationship dictates that the middle term $\binom{m}{2}$ is the average of the first and fifth terms, which is expressed as $2 \binom{m}{2} = \binom{m}{1} + \binom{m}{3}$. Algebraic simplification yields $m(m-1) = m + \frac{m(m-1)(m-2)}{6}$, which leads to $(m-2)(1 - \frac{m-1}{6}) = 0$. Since $m=2$ results \in a trivial expansion, the valid solution is $m=7$.

Substituting $m=7$ into the sixth term of the expansion $(A+B)^7$, defined as $\binom{7}{5} A^{7-5} B^5$ with $A = \sqrt{10-3^x}$ and $B = \sqrt[5]{3^{x-2}}$, yields the equation $21(10-3^x)(3^{x-2}) = 21$. Dividing both sides by 21 and expressing $3^{x-2}$ as $\frac{3^x}{9}$ results \in $(10-3^x) \frac{3^x}{9} = 1$, which rearranges into the quadratic form $(3^x)^2 - 10 \cdot 3^x + 9 = 0$. Factoring this expression gives $(3^x - 9)(3^x - 1) = 0$, leading to $3^x = 9$ or $3^x = 1$. The corresponding values for $x$ are 2 and 0, and the \sum of their squares is $0^2 + 2^2 = 4$.

The sum of the coefficients of odd powers in the binomial expansion $(1+x)^n$ relies on the property that the sum of odd-indexed coefficients equals $2^{n-1}$. Applied to $(1+x)^{99}$, this gives $K = 2^{98}$. For the expansion $(2+\frac{1}{\sqrt{2}})^{200}$, the middle term corresponds to $T_{101}$, calculated as $\binom{200}{100} (2)^{100} (\frac{1}{\sqrt{2}})^{100}$. Simplifying the power of 2 leads to $a = \binom{200}{100} 2^{50}$.

Substituting these results into the ratio $\frac{\binom{200}{99} K}{a}$ produces $\frac{\binom{200}{99} 2^{98}}{\binom{200}{100} 2^{50}}$. The ratio of binomial coefficients $\frac{\binom{200}{99}}{\binom{200}{100}}$ evaluates to $\frac{100}{101}$. Combining this with the powers of 2, the expression becomes $\frac{100}{101} 2^{48}$. Using the identity $100 = 25 \cdot 2^2$, the expression becomes $\frac{25 \cdot 2^{50}}{101}$. Matching this to the form $\frac{2^\ell m}{n}$ reveals $\ell = 50$, $m = 25$, and $n = 101$, establishing the required ordered pair as $(50, 101)$.

The binomial theorem establishes that in an expansion $(a+b)^n$, the terms from the beginning and the end maintain a reciprocal relationship where the $r$-th term from the end is equivalent to the $r$-th term from the beginning if the terms $a$ and $b$ are interchanged. This symmetry allows the binomial coefficients to cancel out when calculating the ratio between the $r$-th term from the beginning and the $r$-th term from the end, simplifying the expression to a power of the ratio of the two terms themselves.

In the expansion $(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})^n$, the fifth term from the beginning is $T_5 = {}^{n}C_4 (2^{1/4})^{n-4} (3^{-1/4})^4$, and the fifth term from the end is $T'_5 = {}^{n}C_4 (3^{-1/4})^{n-4} (2^{1/4})^4$. Dividing these yields the ratio $\frac{T_5}{T'_5} = \frac{2^{(n-4)/4} \cdot 3^{-1}}{3^{-(n-4)/4} \cdot 2^1}$, which simplifies to $2^{(n-8)/4} \cdot 3^{(n-8)/4} = 6^{(n-8)/4}$. Setting this expression equal to the given value of $\sqrt{6}$, or $6^{1/2}$, provides the exponent equation $\frac{n-8}{4} = \frac{1}{2}$, which solves to $n=10$.

With $n=10$, we calculate the third term from the beginning using the general formula $T_3 = {}^{10}C_2 (2^{1/4})^8 (3^{-1/4})^2$. Evaluating the components gives ${}^{10}C_2 = 45$, $(2^{1/4})^8 = 2^2 = 4$, and $(3^{-1/4})^2 = 3^{-1/2} = \frac{1}{\sqrt{3}}$. Multiplying these together results in $45 \cdot 4 \cdot \frac{1}{\sqrt{3}} = \frac{180}{\sqrt{3}}$, which simplifies to $60\sqrt{3}$.

Rationalizing the given expression $\left[ \frac{2}{\sqrt{5x^3+1} - \sqrt{5x^3-1}} \right]^8 + \left[ \frac{2}{\sqrt{5x^3+1} + \sqrt{5x^3-1}} \right]^8$, the terms simplify to $(\sqrt{5x^3+1} + \sqrt{5x^3-1})^8 + (\sqrt{5x^3+1} - \sqrt{5x^3-1})^8$. Let $A = \sqrt{5x^3+1}$ and $B = \sqrt{5x^3-1}$, so the expression becomes $(A+B)^8 + (A-B)^8$, which expands to $2 \left[ \binom{8}{0}A^8 + \binom{8}{2}A^6B^2 + \binom{8}{4}A^4B^4 + \binom{8}{6}A^2B^6 + \binom{8}{8}B^8 \right]$. Since each term $A^2 = 5x^3+1$ and $B^2 = 5x^3-1$ is a polynomial of degree 3, the highest degree of $x$ in the expansion is $x^{12}$, making $n = 12$. The coefficient $m$ is determined by the sum of the coefficients of $x^{12}$ from each binomial term, which results in $m = 80,000 = 8(10)^4$, leading to the ordered pair $(12, 8(10)^4)$.