Binomial Theorem – JEE Main MathematicsPractice Questions & PYQs

An arithmetico-geometric progression is formed by the product of corresponding terms from an arithmetic progression and a geometric progression. Given the third term of the sequence $a_1, a_2, 2, a_3, a_4$ is $2$ and the common ratio of the geometric component is $2$, we can define the five terms by working symmetrically around this middle value. Letting $a = 2$ and $d$ be the common difference of the arithmetic progression, the five terms are expressed as $\frac{a-2d}{4}, \frac{a-d}{2}, a, 2(a+d), 4(a+2d)$.

Substituting $a = 2$ into these expressions, the progression becomes $\frac{2-2d}{4}, \frac{2-d}{2}, 2, 2(2+d), 4(2+2d)$. The sum of these five terms is given as $\frac{49}{2}$, which leads to the following equation:
$\frac{2-2d}{4} + \frac{2-d}{2} + 2 + 4 + 2d + 8 + 8d = \frac{49}{2}$
Simplifying the terms involving $d$ and the constant values results in:
$0.5 - 0.5d + 1 - 0.5d + 14 + 10d = \frac{49}{2}$
$15.5 + 9d = 24.5$
Subtracting $15.5$ from both sides gives $9d = 9$, which simplifies to $d = 1$. The term $a_4$ corresponds to the fifth term in the sequence, $4(2+2d)$. Substituting $d = 1$ yields $4(2+2) = 16$.

The general term in a binomial expansion of the form $(x+y)^n$ is defined by $T_{r+1} = {}^{n}C_r x^{n-r} y^r$, which allows us to isolate specific powers of a variable by solving for $r$. In the first expression $(ax^3 + \frac{1}{bx^{1/3}})^{15}$, the general term is ${}^{15}C_r (ax^3)^{15-r} (b^{-1}x^{-1/3})^r$. To find the coefficient of $x^{15}$, we set the exponent of $x$ equal to 15, resulting in the equation $45 - 3r - \frac{r}{3} = 15$. This simplifies to $30 = \frac{10r}{3}$, giving $r = 9$, which leads to the coefficient ${}^{15}C_9 a^6 b^{-9}$.

In the second expression $(ax^{1/3} - \frac{1}{bx^3})^{15}$, the general term is ${}^{15}C_r (ax^{1/3})^{15-r} (-b^{-1}x^{-3})^r$. Setting the exponent of $x$ equal to -15 provides the equation $\frac{15-r}{3} - 3r = -15$. This simplifies to $5 - \frac{10r}{3} = -15$, or $20 = \frac{10r}{3}$, yielding $r = 6$. The resulting coefficient is ${}^{15}C_6 a^9 (-1)^6 b^{-6}$, which simplifies to ${}^{15}C_6 a^9 b^{-6}$.

Equating these two coefficients gives ${}^{15}C_9 a^6 b^{-9} = {}^{15}C_6 a^9 b^{-6}$. Because the combination terms ${}^{15}C_9$ and ${}^{15}C_6$ are equal, they cancel out. The remaining equation $a^6 b^{-9} = a^9 b^{-6}$ can be rearranged by dividing both sides by $a^6 b^{-6}$, resulting in $b^{-3} = a^3$. This simplifies to $1 = a^3 b^3$, confirming the relationship $ab = 1$.

The general term formula $T_{r+1} = {}^{n}C_r a^{n-r} b^r$ is the foundation for identifying specific coefficients in binomial expansions. For the expansion $(\alpha x^3 + \frac{1}{\beta x})^{11}$, the general term is ${}^{11}C_r (\alpha x^3)^{11-r} (\frac{1}{\beta x})^r$, which simplifies to ${}^{11}C_r \alpha^{11-r} \beta^{-r} x^{33-4r}$; setting the power of $x$ to $9$ requires $33-4r = 9$, which identifies $r=6$ and results in the coefficient ${}^{11}C_6 \frac{\alpha^5}{\beta^6}$. For the second expansion $(\alpha x - \frac{1}{\beta x^3})^{11}$, the general term is ${}^{11}C_r (\alpha x)^{11-r} (-\frac{1}{\beta x^3})^r$, simplifying to ${}^{11}C_r \alpha^{11-r} (-1)^r \beta^{-r} x^{11-4r}$. Setting the power of $x$ to $-9$ requires $11-4r = -9$, which identifies $r=5$, producing the coefficient $-{}^{11}C_5 \frac{\alpha^6}{\beta^5}$. Since these coefficients are equal and the binomial coefficients satisfy ${}^{11}C_6 = {}^{11}C_5$, equating them leads to ${}^{11}C_6 \frac{\alpha^5}{\beta^6} = -{}^{11}C_5 \frac{\alpha^6}{\beta^5}$. Canceling the common combinations leaves $\frac{\alpha^5}{\beta^6} = -\frac{\alpha^6}{\beta^5}$, which simplifies to $\frac{1}{\beta} = -\alpha$. Therefore, $\alpha\beta = -1$, and squaring both sides confirms that $(\alpha\beta)^2 = 1$.

Legendre's formula determines the highest power of a prime $p$ that divides $n!$ by summing the integer quotients of $n$ divided by increasing powers of $p$. This method systematically counts how many times the prime appears as a factor throughout the product of integers from 1 up to $n$.

For $66!$ with $p = 3$, the necessary calculations involve determining the floor of each division: $\lfloor \frac{66}{3} \rfloor = 22$, $\lfloor \frac{66}{9} \rfloor = 7$, and $\lfloor \frac{66}{27} \rfloor = 2$. The process stops here because the next power, 81, is greater than 66. Adding these values, $22 + 7 + 2 = 31$, gives the largest natural number $n$ such that $3^n$ divides $66!$.

Expanding $(1+x)^p$ and $(1-x)^q$ allows us to identify the components that form the overall coefficient of each power of $x$ when these series are multiplied. The linear term emerges from the product of the $x$ terms and constant terms, yielding the equation:
$p-q=4$
For the quadratic term, we combine the product of the $x^2$ terms with the constant terms and the product of the two linear terms, which simplifies to the following relation:
$\frac{p(p-1)}{2}+\frac{q(q-1)}{2}-pq=-5$
By substituting $p=q+4$ into the quadratic equation expressed as $p^2+q^2-p-q-2pq=-10$, we obtain $(q+4)^2+q^2-(q+4)-q-2(4+q)q=-10$. After distributing the values and grouping like terms, the algebra reduces to $-2q=-22$, confirming that $q=11$. This implies $p=15$, and evaluating the target expression $2p+3q$ as $2(15)+3(11)$ results in $63$.

The binomial expansion of $\left( x - \frac{3}{x^2} \right)^n$ is governed by the general term formula ${n \choose r} x^{n-r} \left( -\frac{3}{x^2} \right)^r$. Identifying the integer $n$ requires examining the first three terms of the expansion, which correspond to $r=0, 1, 2$. These terms are ${n \choose 0} x^n$, ${n \choose 1} x^{n-1} \left( -\frac{3}{x^2} \right)$, and ${n \choose 2} x^{n-2} \left( -\frac{3}{x^2} \right)^2$. Equating the sum of their coefficients to 376 leads to the equation $1 - 3n + 9 \frac{n(n-1)}{2} = 376$. Multiplying by 2 and simplifying yields $9n^2 - 15n - 750 = 0$, which reduces to $3n^2 - 5n - 250 = 0$. Factoring gives $(n-10)(3n+25) = 0$, and since $n$ must be a natural number, $n=10$.

With $n$ determined, the general term is ${10 \choose r} x^{10-r} (-3)^r x^{-2r}$, which simplifies to ${10 \choose r} (-3)^r x^{10-3r}$. The coefficient of $x^4$ occurs when the exponent $10-3r = 4$, implying $3r=6$ or $r=2$. Substituting $r=2$ into the expression ${10 \choose r} (-3)^r$ gives ${10 \choose 2} (-3)^2$. Calculating this, ${10 \choose 2}$ is 45 and $(-3)^2$ is 9, so the final coefficient is 405.

To determine the parity of the integer parts of these expressions, we can utilize the properties of algebraic conjugates. For the expression $x = (8\sqrt{3}+13)^{13}$, we introduce its conjugate term $x' = (8\sqrt{3}-13)^{13}$. Since $8\sqrt{3} = \sqrt{192} \approx 13.85$, the value $8\sqrt{3}-13$ is a positive fraction less than $1$, which implies $0 < x' < 1$. When we expand $x$ and $x'$ using the binomial theorem and subtract them, the even-powered terms cancel out, leaving twice the sum of the odd-powered terms: $x - x' = 2 [ \binom{13}{1}(8\sqrt{3})^{12}(13)^1 + \binom{13}{3}(8\sqrt{3})^{10}(13)^3 + \dots ]$. Because every term within the bracket involves an even power of $8\sqrt{3}$-such as $(8\sqrt{3})^{12} = 192^6$-each component is an integer, making $x - x'$ an even integer. Given $x = [x] + f$ where $0 < f < 1$, the relation $x - x' = [x] + f - x'$ simplifies to $[x] = x - x'$ under the condition $f=x'$, confirming that $[x]$ is an even integer.

We apply the exact same reasoning to $y = (7\sqrt{2}+9)^9$ by considering its conjugate $y' = (7\sqrt{2}-9)^9$. With $7\sqrt{2} = \sqrt{98} \approx 9.89$, it is clear that $0 < y' < 1$. The difference $y - y' = 2 [ \binom{9}{1}(7\sqrt{2})^8(9)^1 + \binom{9}{3}(7\sqrt{2})^6(9)^3 + \dots ]$ represents an even integer, as each term in the expansion contains an even power of $7\sqrt{2}$, such as $(7\sqrt{2})^8 = 98^4$. This confirms $[y]$ is also an even integer. Since both $[x]$ and $[y]$ are even, their sum $[x] + [y]$ must be even.

The binomial theorem identifies the general term of an expansion $(a+b)^n$ as $T_{r+1} = {}^{n}C_r a^{n-r} b^r$. Applying this to $\left(x^{2/3} + 2x^{-3}\right)^{30}$, the general term is given by:
$T_{r+1} = {}^{30}C_r \cdot 2^r \cdot x^{\frac{60-11r}{3}}$
To obtain a term of the form $\beta x^{-\alpha}$ with $\alpha > 0$, the exponent must be negative, requiring $\frac{60-11r}{3} = -\alpha$, or $\alpha = \frac{11r-60}{3}$. Since $\beta = {}^{30}C_r 2^r$ is inherently a natural number, we only need to satisfy the inequality $\alpha > 0$, which implies $11r > 60$ or $r > 5.45$. The smallest integer $r$ that satisfies this condition is $r=6$, and substituting this into our equation for $\alpha$ gives $\alpha = \frac{11(6)-60}{3}$, which results in $2$.

The insertion of two arithmetic means $A_1$ and $A_2$ between two positive numbers $a$ and $b$ creates an arithmetic progression where the sum of the means simplifies to $A_1 + A_2 = a + b$. Similarly, inserting three geometric means $G_1, G_2, G_3$ between $a$ and $b$ establishes a geometric progression where the product of the first and third geometric means is $G_1 G_3 = ab$.

Evaluating the expression $G_1^4 + G_2^4 + G_3^4 + G_1^2 G_3^2$ begins by representing each term in terms of $a$ and $b$, where $G_1^4 = a^3b$, $G_2^4 = a^2b^2$, and $G_3^4 = ab^3$. Since $G_1^2 = a^{3/2}b^{1/2}$ and $G_3^2 = a^{1/2}b^{3/2}$, the term $G_1^2 G_3^2$ simplifies to $a^2b^2$. Summing these components yields $a^3b + a^2b^2 + ab^3 + a^2b^2$, which aggregates to $a^3b + 2a^2b^2 + ab^3$. Factoring out the common $ab$ results in $ab(a^2 + 2ab + b^2)$, which is equivalent to $ab(a+b)^2$. Replacing $ab$ with $G_1 G_3$ and $(a+b)$ with $A_1 + A_2$ yields $(A_1 + A_2)^2 G_1 G_3$.

This expression follows the pattern of an arithmetico-geometric series, where each term decreases arithmetically while the denominator grows geometrically. A standard technique to resolve such series is to multiply the entire sum $S$ by the common ratio $\frac{1}{5}$, creating a shifted version of the series that aligns nearly identical terms when subtracted from the original $S$. Performing this subtraction, denoted as $S - \frac{S}{5}$, leaves 109 at the front, followed by a sequence of negative fractional terms that simplifies to $-(\frac{1}{5} + \frac{1}{5^2} + \dots + \frac{1}{5^{109}})$, and concluding with the final term $-\frac{1}{5^{109}}$.

The parenthetical part represents a standard geometric series with 109 terms, summing to $\frac{\frac{1}{5}(1 - \frac{1}{5^{109}})}{1 - \frac{1}{5}}$, which calculates to $\frac{1}{4}(1 - \frac{1}{5^{109}})$. Substituting this into the equation results in $\frac{4S}{5} = 109 - \frac{1}{4} + \frac{1}{4 \cdot 5^{109}}$. By isolating $S$ and multiplying by 16, we can evaluate the required expression. Specifically, $16S = 16(\frac{5}{4})(109 - \frac{1}{4} + \frac{1}{4 \cdot 5^{109}})$, which expands to $20 \cdot 109 - 5 + \frac{1}{5^{108}}$. Since $(25)^{-54}$ is equivalent to $\frac{1}{5^{108}}$, subtracting this term from $16S$ cancels out the fractional component, leaving 2180 - 5, which results in 2175.

A constant term in a binomial expansion exists where the powers of the variable $x$ combine to equal zero, a condition we verify using the general term formula $T_{r+1} = {}^{7}C_{r} (3 x^2)^{7-r} (-\frac{1}{2 x^5})^r$. Simplifying the exponents of $x$ leads to $14 - 7r = 0$, indicating that the constant term corresponds to $r = 2$. Substituting this value into the expression yields $\alpha = {}^{7}C_{2} \cdot 3^5 \cdot (-\frac{1}{2})^2$, which calculates to $\frac{21 \times 243}{4}$, or $1275.75$. Finding the greatest integer less than or equal to this result gives $1275$.

The binomial theorem provides a direct method for identifying any specific term in an expansion using the general term formula $T_{r+1} = {}^{n}C_r a^{n-r} b^r$. For the expression $\left(x^4 - \frac{1}{x^3}\right)^{15}$, the general term is given by ${}^{15}C_r (x^4)^{15-r} (-x^{-3})^r$. Consolidating the variables allows this to be rewritten as ${}^{15}C_r (-1)^r x^{60-4r} x^{-3r}$, which simplifies to ${}^{15}C_r (-1)^r x^{60-7r}$. Because the target is the coefficient of $x^{18}$, equate the exponent $60 - 7r$ to 18, leading to $7r = 42$ and thus $r = 6$. Substituting $r = 6$ into the term gives ${}^{15}C_6 (-1)^6$, which simplifies to the combination ${}^{15}C_6$. Evaluating this combination yields the final coefficient of 5005.

When a sequence exhibits constant second differences, its general term $T_n$ follows a quadratic form $T_n = a n^2 + b n + c$. By observing the terms 4, 11, 21, 34, and 50, the first differences are 7, 10, 13, and 16, while the second differences are a constant 3, confirming the quadratic nature of the sequence. Setting up equations for the first three terms gives $a+b+c=4$, $4a+2b+c=11$, and $9a+3b+c=21$, which yields coefficients $a = \frac{3}{2}$, $b = \frac{5}{2}$, and $c = 0$. Consequently, the general term is $T_n = \frac{3}{2} n^2 + \frac{5}{2} n$.

To determine the sum of $n$ terms $S_n$, we apply the summation formulas $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^n k = \frac{n(n+1)}{2}$. Combining these results with our general term, $S_n = \frac{3}{2} \sum k^2 + \frac{5}{2} \sum k$ simplifies to the expression $S_n = \frac{n(n+1)(n+3)}{2}$. Using this formula, we evaluate the required expression $\frac{1}{60} (S_{29} - S_9)$ by calculating the terms for 29 and 9 respectively. Substituting these into the formula results in $\frac{1}{60} \left( \frac{29 \cdot 30 \cdot 32}{2} - \frac{9 \cdot 10 \cdot 12}{2} \right)$, which simplifies to 223.

The identity below establishes that the difference of two powers, $x^n - y^n$, is always divisible by the difference of the bases, $x - y$:

$x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + \dots + y^{n-1})$

Applying this to the first statement, $2023^{2022} - 1999^{2022}$ must be divisible by $2023 - 1999$, which equals $24$. Since $24$ is a multiple of $8$, the entire expression is necessarily divisible by $8$, confirming the first statement is correct.

The divisibility of $13(1+12)^n - 11n - 13$ by $144$ can be tested by expanding the power term using the binomial theorem:

$13(1 + n(12) + \binom{n}{2}(144) + \dots) - 11n - 13 = 145n + 13\binom{n}{2}(144) + \dots$

This expansion leaves $145n$ combined with a series of terms that are all multiples of $144$. Because the expression becomes a multiple of $144$ whenever $n$ is chosen as a multiple of $144$, such as $n = 144m$ for any positive integer $m$, there are infinitely many such values, confirming the second statement is also correct.

The mathematical identity $\sum_{r=0}^{n} r^2 \binom{n}{r} = n(n+1)2^{n-2}$ provides a direct way to evaluate the required sum. By substituting $n = 2023$ into this expression, we obtain the product $2023 \times 2024 \times 2^{2021}$. Equating this result to the provided expression $2023 \times \alpha \times 2^{2022}$ and dividing both sides by $2023 \times 2^{2021}$ allows us to isolate $\alpha$. This calculation yields $\alpha = 1012$.

Fermat's Little Theorem states that for any prime $p$ and integer $a$ not divisible by $p$, $a^{p-1} \equiv 1 \pm od p$. Applying this to $p=11$ and $a=5$, we establish that $5^{10} \equiv 1 \pm od{11}$, which allows us to simplify $5^{99}$ to $(5^{10})^9 \cdot 5^9 \equiv 1^9 \cdot 5^9 \equiv 5^9 \pm od{11}$. Calculating this power by observing that $5^2 = 25 \equiv 3 \pm od{11}$ and $5^4 \equiv 3^2 = 9 \pm od{11}$, we find $5^9 = 5^4 \cdot 5^4 \cdot 5^1 \equiv 9 \cdot 9 \cdot 5 = 81 \cdot 5 \equiv 4 \cdot 5 = 20 \equiv 9 \pm od{11}$.

The expression $(21-2)^{200} + (21+2)^{200}$ expands using the binomial theorem such that every term containing a factor of $21^2$ or higher becomes divisible by $441$, which is a multiple of $49$. Consequently, these terms vanish under modulo $49$, reducing the entire sum to $2 \times 2^{200}$, or $2^{201}$.

Rewriting $2^{201}$ as $(2^3)^{67}$ yields $8^{67}$, which can be expressed as $(1+7)^{67}$. Expanding this via the binomial theorem results in $1 + 67 \times 7$ plus higher-order terms that are multiples of $7^2$, or $49$. Because $67 \times 7 = 469$ and $469 = 9 \times 49 + 28$, the expression simplifies to $1 + 28$ mod $49$, resulting in a final remainder of $29$.

In the binomial expansion of $\left(3^{\frac{1}{2}}+5^{\frac{1}{4}}\right)^{680}$, the general term is defined by the formula $T_{r+1} = \binom{680}{r} 3^{\frac{680-r}{2}} 5^{\frac{r}{4}}$, where the index $r$ can be any integer from $0$ to $680$. An integral term is produced only when both exponents, $\frac{680-r}{2}$ and $\frac{r}{4}$, are integers.

The requirement that $\frac{r}{4}$ must be an integer dictates that $r$ must be a multiple of $4$, resulting in the possible values $r \in \{0, 4, 8, \dots, 680\}$. Because every multiple of $4$ is inherently an even number, substituting these values into the first exponent $\frac{680-r}{2}$ consistently results in an integer as well, satisfying both conditions for all values in this set.

To determine the total count of these terms, we treat the sequence $0, 4, 8, \dots, 680$ as an arithmetic progression. Applying the formula for the $n$-th term of an arithmetic progression, $a_n = a + (n-1)d$, where $a_n = 680$, $a = 0$, and the common difference $d = 4$, we get $680 = 0 + (n-1)4$. Solving this equation gives $170 = n - 1$, which results in $n = 171$. There are exactly 171 integral terms in this expansion.

The binomial coefficient $\binom{n}{r}$ represents the number of ways to choose $r$ objects from a set of $n$, and is defined by the formula $\frac{n!}{r!(n-r)!}$. When comparing the ratio $\binom{2n}{3}:\binom{n}{3} = 10:1$, we write this as a fraction where the common terms in the expansion cancel out. Since the factor $3!$ exists in both the numerator and denominator, it is eliminated immediately, allowing us to reduce the expression to a simpler algebraic form involving $n$.

By expanding the terms based on the factorial definition, we obtain the equation $\frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)} = 10$. We can factor a 2 out of the term $2n-2$ to see that it equals $2(n-1)$, which then cancels with the $(n-1)$ in the denominator. This simplification leaves us with $\frac{4(2n-1)}{n-2} = 10$. Multiplying both sides by $n-2$ results in $8n - 4 = 10n - 20$, which rearranges to $2n = 16$, identifying that $n = 8$.

With the value of $n$ determined, we substitute it into the required ratio $\frac{n^2+3n}{n^2-3n+4}$. Evaluating the numerator gives $64 + 24 = 88$, and calculating the denominator yields $64 - 24 + 4 = 44$. Dividing these two values, $\frac{88}{44}$, results in a final ratio of $2:1$.

The general term for a binomial expansion $(A + B)^n$ is defined by the formula $T_{r+1} = {}^{n}C_{r} A^{n-r} B^r$. Applying this to the expression $(ax - \frac{1}{bx^2})^{13}$, the general term becomes ${}^{13}C_{r} (ax)^{13-r} (-\frac{1}{bx^2})^r$, which simplifies to ${}^{13}C_{r} a^{13-r} (-1)^r b^{-r} x^{13-3r}$. To find the coefficient of $x^7$, we set the exponent $13-3r$ equal to $7$, leading to $3r = 6$ or $r=2$. Substituting $r=2$ into the expression, the coefficient is ${}^{13}C_2 a^{11} b^{-2}$.

For the second expression $(ax + \frac{1}{bx^2})^{13}$, the general term is ${}^{13}C_{r} (ax)^{13-r} (\frac{1}{bx^2})^r$, which simplifies to ${}^{13}C_{r} a^{13-r} b^{-r} x^{13-3r}$. Setting the exponent $13-3r$ equal to $-5$ gives $3r = 18$, so $r=6$. The coefficient of $x^{-5}$ is therefore ${}^{13}C_6 a^7 b^{-6}$.

Equating these two coefficients, we have ${}^{13}C_2 a^{11} b^{-2} = {}^{13}C_6 a^7 b^{-6}$. Dividing both sides by $a^7$ and multiplying by $b^6$ isolates $a^4 b^4$ on one side of the equation, resulting in $a^4 b^4 = \frac{{}^{13}C_6}{{}^{13}C_2}$. Calculating the binomial coefficients, ${}^{13}C_6$ equals $1716$ and ${}^{13}C_2$ equals $78$. Dividing $1716$ by $78$ yields $22$.