Binomial Theorem – JEE Main MathematicsPractice Questions & PYQs

The expression can be evaluated by splitting it into two distinct parts: the alternating series involving binomial coefficients multiplied by indices, and the sum of specific odd-indexed binomial coefficients. For the first part, we utilize the identity $r \cdot {}^{n}C_r = n \cdot {}^{n-1}C_{r-1}$ to transform the sum into $15 \sum_{r=1}^{15} (-1)^r {}^{14}C_{r-1}$. Expanding this yields $15(-{}^{14}C_0 + {}^{14}C_1 - {}^{14}C_2 + \dots - {}^{14}C_{14})$, which effectively equals $15 \times 0$ because it is a binomial expansion of $(1-1)^{14}$.

The second part consists of the summation ${}^{14}C_1 + {}^{14}C_3 + \dots + {}^{14}C_{11}$. Since the sum of odd-indexed binomial coefficients for a power $n$ is known to be $2^{n-1}$, we recognize that the complete series ${}^{14}C_1 + {}^{14}C_3 + \dots + {}^{14}C_{13}$ equals $2^{13}$. Because the given series stops at ${}^{14}C_{11}$, we subtract the missing term ${}^{14}C_{13}$, which is equivalent to ${}^{14}C_1$ or 14. Combining these findings results in a final value of $0 + 2^{13} - 14$, which simplifies to $2^{13}-14$.

Expanding the binomials $(1-y)^m$ and $(1+y)^n$ allows us to express the product as a power series, where the coefficients are determined by the binomial theorem. The product of the expansions $(1 - my + \frac{m(m-1)}{2}y^2 - \dots)$ and $(1 + ny + \frac{n(n-1)}{2}y^2 + \dots)$ generates the coefficients for the combined series.

The coefficient of $y$, given as $a_1$, emerges from the linear combinations, resulting in $n - m = 10$. Similarly, the coefficient of $y^2$, given as $a_2$, is obtained by summing the products that yield a quadratic term, which results in the following equation:
$\frac{n(n-1)}{2} - mn + \frac{m(m-1)}{2} = 10$
Multiplying this expression by two simplifies it to $n^2 - n + m^2 - m - 2mn = 20$. Grouping the terms reveals the identity $(n - m)^2 - (n + m) = 20$.

Substituting the known value $n - m = 10$ into this expression leads to $10^2 - (n + m) = 20$. Since $10^2$ is 100, subtracting 20 from 100 identifies the value of $(n + m)$ as 80.

The term independent of $t$ occurs when the net exponent of $t$ in the general expansion is zero. Expanding $(t x^{1/5} + t^{-1}(1-x)^{1/10})^{10}$ involves the general term ${}^{10}C_r (t x^{1/5})^{10-r} (t^{-1}(1-x)^{1/10})^r$, where the variable $t$ effectively appears as $t^{10-2r}$. Setting this exponent to zero dictates that $r=5$, defining the independent term as $T_6 = {}^{10}C_5 x(1-x)^{1/2}$.

Determining the maximum value of this term necessitates maximizing $f(x) = x\sqrt{1-x}$ on the interval $(0, 1)$. Taking the derivative and equating it to zero yields $\sqrt{1-x} - \frac{x}{2\sqrt{1-x}} = 0$, which simplifies to $2(1-x) - x = 0$, establishing $x = 2/3$ as the value for the maximum. Substituting this result into the expression for $T_6$ gives ${}^{10}C_5 \cdot \frac{2}{3} \cdot \sqrt{1/3}$, which simplifies to $\frac{2 \cdot 10!}{3\sqrt{3}(5!)^2}$.

The core principle for simplifying this series is the Hockey-stick identity, which establishes that the sum of consecutive binomial coefficients $\sum_{k=r}^{n} {}^{k}C_{r}$ equals ${}^{n+1}C_{r+1}$. Applying this to the summation provided, the sequence ${}^{2}C_{2}+{}^{3}C_{2}+\dots+{}^{n}C_{2}$ collapses into a single term, ${}^{n+1}C_{3}$. Substituting this back into the original expression gives ${}^{n+1}C_{2} + 2({}^{n+1}C_{3})$.

To finalize the simplification, we can decompose the coefficient of 2 into $1+1$, which allows us to regroup the terms as $\left({}^{n+1}C_{2}+{}^{n+1}C_{3}\right) + {}^{n+1}C_{3}$. By utilizing the property ${}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r}$, the first grouping becomes ${}^{n+2}C_{3}$. The expression is now ${}^{n+2}C_{3} + {}^{n+1}C_{3}$. Expanding these binomial coefficients into factorials yields $\frac{(n+2)(n+1)n}{6} + \frac{(n+1)n(n-1)}{6}$. Factoring out the common term $\frac{n(n+1)}{6}$ leaves behind $\frac{n(n+1)}{6} \times [(n+2) + (n-1)]$, which simplifies to the final result of $\frac{n(n+1)(2n+1)}{6}$.

The general $(r+1)$-th term of a binomial $(a + b)^n$ is calculated using the formula $\binom{n}{r} a^{n-r} b^r$. Applying this to the expansion $(x + x^{\log_2 x})^7$ to determine the fourth term involves setting $n=7$ and $r=3$, resulting in $\binom{7}{3} x^4 (x^{\log_2 x})^3$. Since $\binom{7}{3}$ equals 35, the fourth term expression simplifies to $35 \cdot x^{4 + 3\log_2 x}$.

Setting this expression equal to 4480 and dividing by 35 yields $x^{4 + 3\log_2 x} = 128$, which can be rewritten as $x^{4 + 3\log_2 x} = 2^7$. Taking the logarithm base 2 of both sides produces the equation $(4 + 3\log_2 x)\log_2 x = 7$. Substituting $y = \log_2 x$ transforms this into the quadratic equation $3y^2 + 4y - 7 = 0$, which factors as $(3y + 7)(y - 1) = 0$. Given the restriction that $x$ must be a natural number, we discard the negative fractional root and solve $\log_2 x = 1$ to obtain $x = 2$.

The summation $\sum_{r=0}^{29} (30-r) \binom{30}{r}$ relies on two fundamental identities of binomial coefficients: the total sum of coefficients $\sum_{k=0}^{N} \binom{N}{k} = 2^N$ and the property that $\sum_{k=0}^{N} k \binom{N}{k} = N \cdot 2^{N-1}$. Since the term for $r=30$ in the expansion is zero, we can express the entire sum as follows:

$\sum_{r=0}^{30} 30 \binom{30}{r} - \sum_{r=0}^{30} r \binom{30}{r}$

Distributing the summation allows us to evaluate each component separately. The first part simplifies to $30 \cdot 2^{30}$, while the second part equals $30 \cdot 2^{29}$. Subtracting these yields $30 \cdot 2^{30} - 30 \cdot 2^{29}$, which can be factored as $30 \cdot 2^{29} (2 - 1) = 30 \cdot 2^{29}$. Expressing $30$ as $15 \cdot 2$ transforms this value into $15 \cdot 2^{30}$. Comparing this result to $n \cdot 2^{m}$ where $\gcd(2, n)=1$, we determine that $n=15$ and $m=30$, leading to a sum $n+m=45$.

The binomial expansion $(1+x)^{n}$ provides two essential summation identities: $\sum_{k=0}^{n} {}^{n}C_{k} = 2^{n}$ and $\sum_{k=0}^{n} k \cdot {}^{n}C_{k} = n 2^{n-1}$. Setting $n=10$, we distribute the sum $\sum_{k=0}^{10}(4+3k){}^{n}C_{k}$ into $4 \sum_{k=0}^{10} {}^{n}C_{k} + 3 \sum_{k=0}^{10} k \cdot {}^{n}C_{k}$. Applying these identities, the first part becomes $4 \times 2^{10}$ and the second part becomes $3 \times 10 \times 2^{9}$, which simplifies to $15 \times 2^{10}$. Combining these terms yields $4 \times 2^{10} + 15 \times 2^{10} = 19 \times 2^{10}$. Comparing this result to the expression $\alpha \cdot 3^{10} + \beta \cdot 2^{10}$, we find $\alpha = 0$ and $\beta = 19$, resulting in $\alpha + \beta = 19$.

The general term $T_{r+1}$ for the binomial expansion of $(x + y)^n$ is given by ${}^nC_r x^{n-r} y^r$. Applying this structure to the expansion $\left(x + \frac{a}{x^{2}}\right)^{n}$, the general term is expressed as $T_{r+1} = {}^nC_r a^r x^{n-3r}$. The coefficients for the third, fourth, and fifth terms correspond to $r=2, 3,$ and $4$, yielding expressions ${}^nC_2 a^2$, ${}^nC_3 a^3$, and ${}^nC_4 a^4$, respectively.

Using the provided ratio $12:8:3$, we divide the third coefficient by the fourth to get $\frac{{}^nC_2 a^2}{{}^nC_3 a^3} = \frac{3}{a(n-2)} = \frac{12}{8} = \frac{3}{2}$, which simplifies to the relation $a(n-2) = 2$. Similarly, dividing the fourth coefficient by the fifth gives $\frac{{}^nC_3 a^3}{{}^nC_4 a^4} = \frac{4}{a(n-3)} = \frac{8}{3}$, resulting in the relation $a(n-3) = \frac{3}{2}$. Solving this system of equations by subtracting them yields $a(n-2) - a(n-3) = 0.5$, which simplifies to $a = 0.5$, and substituting this into the first equation provides $n = 6$.

The term independent of $x$ occurs when the power of $x$ is zero, satisfying $n-3r = 0$. Since $n=6$, this occurs at $r=2$, identifying $T_3$ as the independent term. Calculating this term results in ${}^6C_2 \cdot \left(\frac{1}{2}\right)^{2} = 15 \cdot \frac{1}{4} = 3.75$, which is equal to $4$.

The term independent of $x$ in a binomial expansion occurs when the powers of $x$ in the general term formula balance to zero. For the expansion $\left(\frac{x}{4} - \frac{12}{x^{2}}\right)^{12}$, the general term is given by $T_{r+1} = \binom{12}{r} \left(\frac{x}{4}\right)^{12-r} \left(-\frac{12}{x^{2}}\right)^{r}$. By isolating the variable components, the exponent of $x$ is $(12-r) - 2r$, or $12-3r$, which must equal $0$ to be independent of $x$, identifying $r=4$ as the required index.

Substituting $r=4$ into the expression yields $T_5 = \binom{12}{4} \left(\frac{1}{4}\right)^{8} (-12)^4$. Since $(-1)^4 = 1$, we can simplify the numerical coefficient to $\binom{12}{4} \cdot \frac{12^4}{4^8}$. Expanding $12^4$ as $(3 \cdot 4)^4$, we obtain $\binom{12}{4} \cdot \frac{3^4 \cdot 4^4}{4^8}$, which reduces to $495 \cdot \frac{3^4}{4^4}$. Equating this to the target form $\left(\frac{3^{6}}{4^{4}}\right) k$, we have $495 \cdot \frac{3^4}{4^4} = \frac{3^6}{4^4} k$, which simplifies to $495 = 3^2 \cdot k$, ultimately giving $k=55$.

The binomial theorem is an essential tool for evaluating remainders when exponents are large, as it allows us to break down bases into forms that are easily divisible by the divisor. We can express $7$ as $(6+1)$ and $10$ as $(9+1)$, expanding the powers $(6+1)^{22}$ and $(9+1)^{22}$ into a series where most terms are multiples of 18. Specifically, these binomial expansions allow us to write the terms as $(1 + 6\lambda)$ and $(1 + 9\mu)$, where $\lambda$ and $\mu$ represent integers containing the higher-order powers of 6 and 9 that are inherently divisible by 18.

Substituting these into the given expression $3(1+6\lambda) + 2(1+9\mu) - 44$, we simplify the constant terms and group the remaining multiples of 18 together. This calculation yields $3 + 2 - 44$, which simplifies to $-39$. Since the remainder must be a positive value in modular arithmetic, we add multiples of 18 until we reach a positive result, specifically $-39 + 54 = 15$. Consequently, the final remainder is 15.

The summation relies on decomposing the general term $(2r+1) \cdot {}^nC_r$ into two series, which leads to $2 \sum_{r=0}^{n} r \cdot {}^nC_r + \sum_{r=0}^{n} {}^nC_r$. Using standard binomial identities, this simplifies to $2(n \cdot 2^{n-1}) + 2^n$, yielding the following result:

$2^n(n+1) = 2^{100} \cdot 101$

From this equivalence, it is clear that $n=100$. Calculating $2[\frac{100-1}{2}]$ gives $2[49.5]$, which evaluates to 98.

The sum of squares of binomial coefficients, expressed as $\sum_{k=0}^{n} ({}^{n}C_k)^2$, follows the identity ${}^{2n}C_n$. This formula arises because ${}^{n}C_k$ is equal to ${}^{n}C_{n-k}$, allowing the sum to be interpreted as the coefficient of $x^n$ in the polynomial product $(1+x)^n(1+x)^n$, which expands to $(1+x)^{2n}$. By substituting $n=20$ into this identity, the expression simplifies directly to ${}^{40}C_{20}$.

To determine how many values of $n$ satisfy the inequality $11^n > 10^n + 9^n$, we can evaluate the terms for initial integer values. For $n=1$, the expression $11 > 19$ is false, for $n=2$ the relation $121 > 181$ is false, for $n=3$ the comparison $1331 > 1729$ fails, and for $n=4$ the inequality $14641 > 16561$ is also false.

However, for $n=5$, we find that $161051 > 159049$, which is true. Since the exponential growth of the left side dominates the sum on the right for all larger values, the inequality holds for every integer from 5 to 100. Subtracting the 4 initial values that do not satisfy the condition from the total count of 100 leaves 96 elements.

The expression $\left(1+\frac{1}{n}\right)^n$ relates to the definition of the mathematical constant $e$, which is represented by the infinite series $1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots$. By applying the binomial theorem to $\left(1+\frac{1}{10^{100}}\right)^{10^{100}}$, we generate a series that starts with $1 + 1 + \frac{1}{2!} + \dots$, where each subsequent term is diminished by factors slightly smaller than one, such as $(1-\frac{1}{n})$. Because these terms are smaller than those in the expansion of $e$ but still yield a sum greater than 2, the total value lies strictly within the interval $(2, 3)$. Given this range, the lowest integer greater than the expression is 3.

To find the remainder of $(2021)^{2023}$ divided by 7, we use modular arithmetic to simplify the base. Since $2021 = 7 \times 288 + 5$, the base simplifies to 5, meaning $(2021)^{2023} \equiv 5^{2023} \pm od 7$. Observing the powers of 5, we find that $5^3 = 125$, and because $125 = 7 \times 18 - 1$, it follows that $5^3 \equiv -1 \pm od 7$. By splitting the exponent 2023 into $3 \times 674 + 1$, the expression becomes $(5^3)^{674} \cdot 5^1$. Substituting the congruence yields $(-1)^{674} \cdot 5$, which simplifies to $1 \cdot 5$, resulting in a final remainder of 5.

Binomial expansions are analyzed using the general term formula $T_{r+1} = \binom{n}{r} a^{n-r} y^r$. For the expression $(x^2 + \frac{1}{bx})^{11}$, the general term is $\binom{11}{r} (x^2)^{11-r} (b^{-1} x^{-1})^r$, which simplifies to $\binom{11}{r} b^{-r} x^{22-3r}$. By setting the power of $x$ equal to $7$, we find $22-3r = 7$, which gives $r=5$, resulting in a coefficient of $\binom{11}{5} b^{-5}$.

The second expression $(x - \frac{1}{bx^2})^{11}$ follows a similar process where the general term is $\binom{11}{k} x^{11-k} (-b^{-1} x^{-2})^k$, simplifying to $\binom{11}{k} (-1)^k b^{-k} x^{11-3k}$. To isolate the coefficient of $x^{-7}$, we set the exponent $11-3k = -7$, yielding $k=6$, which produces a coefficient of $\binom{11}{6} b^{-6}$ since $(-1)^6 = 1$. Equating the two expressions $\binom{11}{5} b^{-5} = \binom{11}{6} b^{-6}$ and noting that $\binom{11}{5} = \binom{11}{6}$ allows us to simplify the relationship to $b^{-5} = b^{-6}$, or $1 = \frac{1}{b}$, confirming that $b=1$.

The binomial expansion of $\left(2 x^{r} + \frac{1}{x^{2}}\right)^{10}$ relies on the general term expression ${}^{10}C_{R} (2 x^{r})^{10-R} (x^{-2})^{R}$, which simplifies to ${}^{10}C_{R} \cdot 2^{10-R} \cdot x^{r(10-R) - 2R}$. For the term to be constant, the total exponent of $x$ must be zero, requiring $r(10-R) - 2R = 0$, or $r = \frac{2R}{10-R}$. The coefficient of this constant term is given as 180, so we must satisfy the equation ${}^{10}C_{R} \cdot 2^{10-R} = 180$. Testing valid integer values for $R$ in the binomial coefficient, we find that when $R=8$, the expression yields ${}^{10}C_{8} \cdot 2^{10-8} = 45 \cdot 4 = 180$. Substituting this value of $R$ into the derived equation for $r$, we calculate $r = \frac{2(8)}{10-8} = \frac{16}{2} = 8$.

Recognizing that the expression for $A$ relies on the binomial theorem expansion $(1-x)^n = \sum_{k=0}^{n} (-1)^k {}^{n}C_{k} x^k$ allows us to simplify each component within the brackets. Applying this theorem individually to the five terms in the sum, where $x$ takes the values $\frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \frac{15}{16}, \text{ and } \frac{31}{32}$, transforms the expression into a sum of geometric powers. Specifically, these terms simplify to $\left(\frac{1}{2}\right)^n + \left(\frac{1}{4}\right)^n + \left(\frac{1}{8}\right)^n + \left(\frac{1}{16}\right)^n + \left(\frac{1}{32}\right)^n$, which can be written as the sum of powers of $\frac{1}{2^n}$.

This sequence represents a finite geometric series with a first term of $\frac{1}{2^n}$, a common ratio of $\frac{1}{2^n}$, and five terms. Utilizing the standard sum formula for a geometric progression, $S = a \frac{1-r^m}{1-r}$, yields:
$A = \frac{\frac{1}{2^n} \left(1 - (\frac{1}{2^n})^5\right)}{1 - \frac{1}{2^n}} = \frac{2^{5n}-1}{2^{5n}(2^n-1)}$
Given the equality $63A = 1 - \frac{1}{2^{30}}$, substituting the derived expression for $A$ results in $63 \left( \frac{2^{5n}-1}{2^{5n}(2^n-1)} \right) = \frac{2^{30}-1}{2^{30}}$. Observing that $5n$ must equal $30$ leads to $n=6$, which satisfies the equation because $2^6 - 1 = 63$, allowing the factor of $63$ to cancel perfectly on the left side.

The binomial expansion theorem determines the general term $T_{r+1}$ of $(a+b)^n$ as $\binom{n}{r} a^{n-r} b^r$. Identifying $n=10$ and $r=8$ for the ninth term, the expression becomes $\binom{10}{8} (25^{x-1}+7)(5^{x-1}+1)^{-1} = 180$, which simplifies to the equation $\frac{25^{x-1}+7}{5^{x-1}+1} = 4$. Substituting $t = 5^{x-1}$ results in the quadratic $t^2 - 4t + 3 = 0$, with roots $t=1$ and $t=3$. Equating $5^{x-1} = 1$ leads to $x-1 = 0$, which yields $x = 1$.

To identify the rational terms in the expansion of $(2^{\frac{1}{3}} + 3^{\frac{1}{4}})^{12}$, we examine the general term given by:
$T_{r+1} = \binom{12}{r} (2^{\frac{1}{3}})^r (3^{\frac{1}{4}})^{12-r}$
For a term to be rational, the exponents of both base numbers must be integers. This requires $\frac{r}{3}$ and $\frac{12-r}{4}$ to be integers, which implies $r$ must be a multiple of 3 and $12-r$ must be a multiple of 4. Considering the range $0 \le r \le 12$, the values for $r$ satisfying the first condition are $\{0, 3, 6, 9, 12\}$ and the values satisfying the second condition are $\{0, 4, 8, 12\}$. The only values that satisfy both conditions simultaneously are $r=0$ and $r=12$.

Substituting these values into the expression allows us to calculate the specific rational terms. When $r=0$, the term is $\binom{12}{0} (2^{\frac{1}{3}})^0 (3^{\frac{1}{4}})^{12} = 1 \cdot 1 \cdot 27 = 27$. When $r=12$, the term is $\binom{12}{12} (2^{\frac{1}{3}})^{12} (3^{\frac{1}{4}})^0 = 1 \cdot 16 \cdot 1 = 16$. The sum of all rational terms is therefore $27 + 16 = 43$.