Binomial Theorem – JEE Main MathematicsPractice Questions & PYQs

The binomial expansion of $(1-2\sqrt{x})^{50}$ generates terms of the form ${}^{50}C_r (1)^{50-r} (-2\sqrt{x})^r$, which simplifies to ${}^{50}C_r (-2)^r x^{r/2}$. For the power of $x$ to be an integer, the exponent $r/2$ must be an integer, which requires the index $r$ to be an even number. Since $r$ is even, the term $(-2)^r$ is equivalent to $2^r$, meaning we need to find the sum of the coefficients ${}^{50}C_r 2^r$ for all even values of $r$ ranging from $0$ to $50$.

This summation corresponds to the sum of even-indexed terms in the expansion of $(1+y)^{50}$ where $y=2$. Leveraging the property that the sum of even-indexed coefficients is given by $\frac{(1+y)^n + (1-y)^n}{2}$, we substitute $n=50$ and $y=2$ into the expression. This yields $\frac{(1+2)^{50} + (1-2)^{50}}{2}$, which further simplifies to $\frac{3^{50} + (-1)^{50}}{2}$ and finally results in $\frac{1}{2}(3^{50}+1)$.

The expansion of the expression $(1+ax+bx^2)(1-2x)^{18}$ involves multiplying $(1+ax+bx^2)$ by the binomial series $\sum_{k=0}^{18} \binom{18}{k}(-2x)^k$. To find the coefficients of specific powers of $x$, we identify the terms that contribute to those powers: $1$ is multiplied by the term $x^n$ from the expansion, $ax$ is multiplied by the term $x^{n-1}$, and $bx^2$ is multiplied by the term $x^{n-2}$.

To determine the coefficient of $x^3$, we sum these contributions and set them to zero: $\binom{18}{3}(-2)^3 + a\binom{18}{2}(-2)^2 + b\binom{18}{1}(-2)^1 = 0$. Substituting the values of the binomial coefficients, we get $-8(816) + 4a(153) - 36b = 0$, which simplifies to $612a - 36b = 6528$, or more cleanly, $51a - 3b = 544$.

For the coefficient of $x^4$, we perform a similar procedure by summing the products contributing to $x^4$, resulting in the equation $\binom{18}{4}(-2)^4 + a\binom{18}{3}(-2)^3 + b\binom{18}{2}(-2)^2 = 0$. Substituting the respective values gives $16(3060) - 8a(816) + 4b(153) = 0$, which simplifies to $48960 - 6528a + 612b = 0$, or $4080 - 544a + 51b = 0$.

Solving the resulting system of linear equations by substituting $b = 17a - \frac{544}{3}$ into the second equation allows us to isolate $a$. Calculating the result yields $a = 16$. Substituting this value back into our first expression gives $b = \frac{272}{3}$, meaning the required pair is $(a,b) = \left(16,\frac{272}{3}\right)$.

The expression inside the parenthesis simplifies significantly if you recognize standard algebraic patterns. Treating $x+1$ as a sum of cubes in the form $(x^{1/3})^3+1^3$, which factors into $(x^{1/3}+1)(x^{2/3}-x^{1/3}+1)$, allows the first fraction to reduce to $x^{1/3}+1$. Similarly, the second fraction involves the difference of squares, where $\frac{x-1}{x-x^{1/2}}$ simplifies to $\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)}$, resulting in $1+x^{-1/2}$. Subtracting these simplified forms leaves $(x^{1/3}-x^{-1/2})^{10}$. Finding the term independent of $x$ involves setting the exponent of $x$ in the general term formula $T_{r+1} = \binom{10}{r} (x^{1/3})^{10-r} (-x^{-1/2})^r$ to zero, which is equivalent to solving $\frac{10-r}{3} - \frac{r}{2} = 0$. Solving this equation for $r$ yields $r=4$, and the final coefficient is calculated as $\binom{10}{4}$, which results in 210.

Expanding both binomial expressions reveals that all terms with even powers of $\sqrt{3}$ cancel out, leaving behind twice the sum of terms with odd powers of $\sqrt{3}$. This simplification yields:

$2 \left[ \binom{2n}{1}(\sqrt{3})^{2n-1} + \binom{2n}{3}(\sqrt{3})^{2n-3} + \dots + \binom{2n}{2n-1}(\sqrt{3})^1 \right]$

Because every term within the bracket involves an odd power of $\sqrt{3}$, the final expression retains an irrational component for any positive integer $n$.

Recognizing the structure of the expression $(1-x-x^2+x^3)^6$ allows for simplification by factoring out common terms, which reveals the product of two binomial expansions. By grouping the terms as $(1-x) - x^2(1-x)$, we rewrite the original expression as $(1-x)^6(1-x^2)^6$, enabling the use of the binomial theorem on each factor independently.

Finding the coefficient of $x^7$ involves identifying pairs of powers $x^a$ from the first expansion and $x^{2b}$ from the second such that $a + 2b = 7$. By multiplying the corresponding binomial terms $(-6x) \cdot (-20x^6)$, $(-20x^3) \cdot (15x^4)$, and $(-6x^5) \cdot (-6x^2)$, we arrive at the following calculation:
$120 - 300 + 36 = -144$

The fundamental identity $\sum_{j=0}^{n} \binom{n}{j} x^j = (1+x)^n$ allows us to evaluate sums involving binomial coefficients by differentiating the polynomial expression with respect to $x$. Applying this method by differentiating once and setting $x=1$ yields the identity $\sum_{j=0}^{n} j \binom{n}{j} = n 2^{n-1}$, whereas a second differentiation supports the evaluation of $\sum_{j=0}^{n} j^2 \binom{n}{j} = n(n+1) 2^{n-2}$. Substituting $n=10$ into these derived formulas, the sum $s_3 = \sum_{j=1}^{10} j^2 \binom{10}{j}$ becomes $10 \times 11 \times 2^8$, which simplifies to $55 \times 2^9$, proving that Statement-1 is true. Regarding Statement-2, the sum $s_1 = \sum_{j=1}^{10} j(j-1) \binom{10}{j}$ calculates to $n(n-1) 2^{n-2} = 90 \times 2^8$. However, the sum $s_2$ correctly evaluates to $10 \times 2^9$, which differs from the $10 \times 2^8$ value claimed in the statement, meaning Statement-2 is false.

The remainder of $8^{2n} - 62^{2n+1}$ when divided by 9 can be determined by evaluating the expression modulo 9, leveraging the property that remainders are preserved under exponentiation and subtraction. Since $8 \equiv -1 \pm od 9$, it follows that $8^{2n} = (8^2)^n = 64^n \equiv 1^n \equiv 1 \pm od 9$. Furthermore, because 63 is a multiple of 9, we have $62 \equiv -1 \pm od 9$, and raising this to the odd power of $2n+1$ results in $62^{2n+1} \equiv (-1)^{2n+1} \equiv -1 \pm od 9$. Substituting these results into the original expression gives $1 - (-1) = 2$, which confirms that the remainder is 2.

The expansion of $\sum_{r=0}^{n} (r+1) {}^n C_r x^r$ is evaluated by splitting the sum into two distinct components: $\sum_{r=0}^{n} r \cdot {}^n C_r x^r$ and $\sum_{r=0}^{n} {}^n C_r x^r$. The second part represents the standard binomial expansion $(1+x)^n$. For the first part, applying the property $r \cdot {}^n C_r = n \cdot {}^{n-1} C_{r-1}$ allows the summation to be rewritten as $n \sum_{r=1}^{n} {}^{n-1} C_{r-1} x^r$, which simplifies to $nx(1+x)^{n-1}$. Combining these expressions yields $(1+x)^n + n x (1+x)^{n-1}$, which confirms that Statement 2 is true.

Setting $x=1$ in this generalized identity directly verifies Statement 1. This substitution results in $(1+1)^n + n(1)(1+1)^{n-1}$, which simplifies to $2^n + n \cdot 2^{n-1}$. Factoring out $2^{n-1}$ produces $2^{n-1}(2+n)$, matching the form provided in Statement 1 and demonstrating that the principle established in Statement 2 serves as a correct explanation for Statement 1.

The general term in the binomial expansion of $(a+x)^n$ is given by $T_{r+1} = {}^nC_r a^{n-r} x^r$. Applying this principle to the binomial $(a-b)^n$, the fifth term ($t_5$) corresponds to $r=4$ and the sixth term ($t_6$) corresponds to $r=5$, resulting in $t_5 = {}^nC_4 a^{n-4} (-b)^4 = {}^nC_4 a^{n-4} b^4$ and $t_6 = {}^nC_5 a^{n-5} (-b)^5 = -{}^nC_5 a^{n-5} b^5$.

Because the sum of these terms is zero, we establish the equality:
${}^nC_4 a^{n-4} b^4 = {}^nC_5 a^{n-5} b^5$
Rearranging the equation to isolate the ratio $\frac{a}{b}$, we divide both sides by $a^{n-5} b^4$:
$\frac{a}{b} = \frac{{}^nC_5}{{}^nC_4}$
By substituting the factorial formulas for the combinations, the ratio simplifies as follows:
$\frac{a}{b} = \frac{\frac{n!}{5!(n-5)!}}{\frac{n!}{4!(n-4)!}} = \frac{4!(n-4)!}{5!(n-5)!} = \frac{4!(n-4)(n-5)!}{5 \cdot 4!(n-5)!} = \frac{n-4}{5}$

The alternating sum ${}^{20}C_0 - {}^{20}C_1 + {}^{20}C_2 - \cdot s + {}^{20}C_{20}$ equals zero because it represents the binomial expansion of $(1-1)^{20}$. Leveraging the symmetry property ${}^{20}C_r = {}^{20}C_{20-r}$, we can observe that the coefficients effectively pair and cancel each other out when the series is partitioned, revealing the role of the central term. By rearranging the equation to isolate the sum up to the tenth index, the calculation simplifies to exactly half of that central term, resulting in $\frac{1}{2} {}^{20}C_{10}$.

Expanding the binomial expressions $(1-y)^m$ and $(1+y)^n$ yields the series $1 - my + \frac{m(m-1)}{2}y^2 - \dots$ and $1 + ny + \frac{n(n-1)}{2}y^2 + \dots$. Multiplying these expansions, the coefficient $a_1$ of the $y$ term is $n - m$, which equals $10$, allowing us to establish the relationship $n = m + 10$. The coefficient $a_2$ of the $y^2$ term is formed by combining $\frac{n(n-1)}{2} - nm + \frac{m(m-1)}{2}$, which also equals $10$.

Multiplying this second equation by $2$ gives $n(n-1) - 2nm + m(m-1) = 20$. Substituting $n = m + 10$ into this expression leads to $(m+10)(m+9) - 2(m+10)m + m(m-1) = 20$. Expanding the terms results in $(m^2 + 19m + 90) - (2m^2 + 20m) + (m^2 - m) = 20$, which simplifies to $-2m + 90 = 20$. Solving this gives $2m = 70$, identifying $m = 35$, which means $n = 45$.

Treating the function as a product of two infinite geometric series, $(1-ax)^{-1}$ and $(1-bx)^{-1}$, allows for the expansion of each as $(1+ax+a^2x^2+\dots+a^nx^n+\dots)$ and $(1+bx+b^2x^2+\dots+b^nx^n+\dots)$. The coefficient of $x^n$ in the resulting product is determined by summing all combinations of products $a^k b^{n-k}$ where the exponents add up to $n$, which generates the sequence $a^n + a^{n-1}b + a^{n-2}b^2 + \dots + b^n$. This sequence is a finite geometric progression consisting of $n+1$ terms, where the first term is $a^n$ and the common ratio is $\frac{b}{a}$. Applying the standard sum formula for a geometric series, $a^n \left[\frac{(b/a)^{n+1}-1}{(b/a)-1}\right]$, leads to the algebraic simplification $\frac{b^{n+1}-a^{n+1}}{b-a}$.

Binomial expansion provides the standard approach for approximating expressions when the variable x is very small, allowing us to drop terms of $x^3$ and higher. We expand each term in the numerator using the general binomial series $(1+u)^n \approx 1 + \nu + \frac{n(n-1)}{2}u^2$. For the first term $(1+x)^{\frac{3}{2}}$, we get $1 + \frac{3}{2}x + \frac{3}{8}x^2$, and for the second term $(1+\frac{1}{2}x)^3$, we expand to $1 + 3(\frac{1}{2}x) + 3(\frac{1}{2}x)^2$, resulting in $1 + \frac{3}{2}x + \frac{3}{4}x^2$. Subtracting the second expansion from the first yields $(\frac{3}{8} - \frac{3}{4})x^2$, which simplifies to $-\frac{3}{8}x^2$. The denominator $(1-x)^{\frac{1}{2}}$ is effectively 1 because any higher-order adjustments from its reciprocal $(1-x)^{-\frac{1}{2}} \approx 1 + \frac{1}{2}x$ would involve multiplying by $-\frac{3}{8}x^2$, producing terms of $x^3$ that are excluded by the problem constraints. Consequently, the entire expression reduces to $-\frac{3}{8}x^2$.

When coefficients of three consecutive terms in a binomial expansion form an arithmetic progression, the middle coefficient must be the arithmetic mean of its neighbors, expressed as:
$2 \cdot {}^mC_r = {}^mC_{r-1} + {}^mC_{r+1}$
For the specific case $r=2$, the coefficients correspond to the $2^{\text{nd}}$, $3^{\text{rd}}$, and $4^{\text{th}}$ terms, which are ${}^mC_1$, ${}^mC_2$, and ${}^mC_3$, and applying the arithmetic progression condition results in the quadratic identity $m^2 - 9m + 14 = 0$. Substituting $r=2$ into the equation $m^2 - m(4r + 1) + 4r^2 - 2 = 0$ yields this identical quadratic form, which verifies the relationship.

The core of this problem relies on Pascal's Identity, which states that ${}^n C_r + {}^n C_{r-1} = {}^{n+1} C_r$. This fundamental identity allows us to merge two combinations with the same upper index and consecutive lower indices into a single term by incrementing the upper index and selecting the higher lower index.

To evaluate the given expression, first expand the summation $\sum_{r=1}^{6} {}^{56-r} C_3$ into its six constituent terms: ${}^{55} C_3 + {}^{54} C_3 + {}^{53} C_3 + {}^{52} C_3 + {}^{51} C_3 + {}^{50} C_3$. When combined with the initial term ${}^{50} C_4$, the entire expression is arranged as ${}^{50} C_4 + {}^{50} C_3 + {}^{51} C_3 + {}^{52} C_3 + {}^{53} C_3 + {}^{54} C_3 + {}^{55} C_3$. Applying the identity recursively from the left, ${}^{50} C_4 + {}^{50} C_3$ simplifies to ${}^{51} C_4$. This result is then added to the next term, ${}^{51} C_3$, yielding ${}^{52} C_4$. This cascading process continues throughout the series, where ${}^{52} C_4 + {}^{52} C_3$ becomes ${}^{53} C_4$, ${}^{53} C_4 + {}^{53} C_3$ becomes ${}^{54} C_4$, and ${}^{54} C_4 + {}^{54} C_3$ becomes ${}^{55} C_4$. The final operation, ${}^{55} C_4 + {}^{55} C_3$, concludes the sequence to result in ${}^{56} C_4$.

The binomial theorem provides a general term formula, ${}^{n}C_r A^{n-r} B^r$, to determine specific coefficients within an expansion $(A+B)^n$. For the first expression $\left[ a x^2 + \frac{1}{b x} \right]^{11}$, the general term is ${}^{11}C_r (a x^2)^{11-r} (b^{-1} x^{-1})^r$, which simplifies to ${}^{11}C_r a^{11-r} b^{-r} x^{22-3r}$. Setting the exponent $22-3r$ equal to $7$ identifies $r=5$ as the required term, yielding a coefficient of ${}^{11}C_5 a^6 b^{-5}$.

For the second expression $\left[ a x - \frac{1}{b x^2} \right]^{11}$, the general term is ${}^{11}C_r (a x)^{11-r} (-b^{-1} x^{-2})^r$, which simplifies to ${}^{11}C_r a^{11-r} (-1)^r b^{-r} x^{11-3r}$. We require the coefficient of $x^{-7}$, so we solve $11-3r = -7$ to find $r=6$, resulting in a coefficient of ${}^{11}C_6 a^5 (-1)^6 b^{-6}$, or simply ${}^{11}C_6 a^5 b^{-6}$. Equating the two coefficients gives ${}^{11}C_5 a^6 b^{-5} = {}^{11}C_6 a^5 b^{-6}$. Since ${}^{11}C_5 = {}^{11}C_6$, we can cancel the combination factors and simplify the powers of $a$ and $b$ to arrive at $a b = 1$.

In a binomial expansion of the form $(1+ax)^n$, the middle term exists when $n$ is even and is located at position $\frac{n}{2} + 1$. For the expression $(1+\alpha x)^4$, the middle term is the third term, which is calculated as ${}^4C_2 (\alpha x)^2$, resulting in a coefficient of $6\alpha^2$. Expanding $(1-\alpha x)^6$ similarly places the middle term at the fourth position, calculated as ${}^6C_3 (-\alpha x)^3$, which yields a coefficient of $-20\alpha^3$. Setting these two coefficients equal to one another requires solving the equation $6\alpha^2 = -20\alpha^3$. By assuming $\alpha$ is non-zero and dividing both sides by $2\alpha^2$, the relationship simplifies to $3 = -10\alpha$, which produces the value $\alpha = -\frac{3}{10}$.

Expanding the expression $(1+x)(1-x)^n$ into $(1-x)^n + x(1-x)^n$ allows us to analyze each part separately to find the coefficient of $x^n$. The general term of $(1-x)^n$ is ${}^n C_r (-1)^r x^r$, which provides a coefficient of ${}^n C_n (-1)^n$ when $r$ equals $n$. For the second part $x(1-x)^n$, the general term becomes ${}^n C_r (-1)^r x^{r+1}$, meaning the coefficient for $x^n$ occurs when $r$ equals $n-1$, yielding ${}^n C_{n-1} (-1)^{n-1}$. Summing these components gives ${}^n C_n (-1)^n + {}^n C_{n-1} (-1)^{n-1}$, and applying the values ${}^n C_n = 1$ and ${}^n C_{n-1} = n$ results in $(-1)^n + n(-1)^{n-1}$, which simplifies to $(-1)^n(1-n)$.

The symmetry property of binomial coefficients, which states that ${}^nC_r = {}^nC_{n-r}$, provides the most direct way to evaluate these summations. We consider $t_n = \sum_{r=0}^n \frac{r}{{}^nC_r}$ and rewrite it by replacing the index $r$ with $n-r$, which yields $t_n = \sum_{r=0}^n \frac{n-r}{{}^nC_{n-r}}$. Since ${}^nC_{n-r}$ is equivalent to ${}^nC_r$, this transformation allows us to write the expression as $t_n = \sum_{r=0}^n \frac{n-r}{{}^nC_r}$.

Adding these two distinct representations of $t_n$ gives $2t_n = \sum_{r=0}^n \frac{r}{{}^nC_r} + \sum_{r=0}^n \frac{n-r}{{}^nC_r}$. Combining the terms under a single summation results in $2t_n = \sum_{r=0}^n \frac{r + n - r}{{}^nC_r}$, which simplifies to $2t_n = \sum_{r=0}^n \frac{n}{{}^nC_r}$. By factoring out the constant $n$, we obtain $2t_n = n \sum_{r=0}^n \frac{1}{{}^nC_r}$. Because the summation term is exactly defined as $S_n$, we arrive at $2t_n = n S_n$, which confirms that the ratio $\frac{t_n}{S_n}$ is $\frac{n}{2}$.

The Binomial Theorem identifies the general term in an expansion $(a+b)^n$ as $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. Applying this to the expression $(\sqrt{3} + \sqrt[8]{5})^{256}$, the general term becomes $\binom{256}{r} (3)^{\frac{256-r}{2}} (5)^{\frac{r}{8}}$, with $r$ taking any integer value from 0 to 256. For this term to be an integer, the exponents must result in whole numbers, meaning $\frac{256-r}{2}$ and $\frac{r}{8}$ must both be integers. The first condition implies that $r$ must be an even number, while the second requires $r$ to be a multiple of 8. Since any multiple of 8 is necessarily an even number, the integral terms occur precisely when $r$ is a multiple of 8. Identifying these values as $0, 8, 16, \dots, 256$ forms an arithmetic progression where the terms are defined by $r = 8k$, with $k$ ranging from 0 up to 32. This calculation yields 33 distinct values for $r$, confirming there are 33 integral terms in the expansion.