Binomial Theorem – JEE Main MathematicsPractice Questions & PYQs

The constant term in a binomial expansion arises when the net exponent of the variable $x$ is zero. For the expression $(\sqrt{x}-\frac{k}{x^{2}})^{10}$, we identify the general term using the formula $T_{r+1} = {}^{10}C_r(\sqrt{x})^{10-r}(-\frac{k}{x^{2}})^r$. Simplifying the powers of $x$ gives $x^{\frac{10-r}{2}} \cdot x^{-2r}$, which combines to $x^{\frac{10-5r}{2}}$. By setting this exponent equal to zero, we find $\frac{10-5r}{2} = 0$, which simplifies to $10 - 5r = 0$, resulting in $r=2$.

Substituting $r=2$ into the general formula provides the constant term $T_3 = {}^{10}C_2(-k)^2$. The value of the binomial coefficient ${}^{10}C_2$ is $45$, leading to the expression $45k^2$. Equating this to the provided value of $405$, we obtain $45k^2 = 405$. Dividing both sides by $45$ gives $k^2 = 9$, and taking the square root confirms that $|k| = 3$.

The coefficients of consecutive terms in a binomial expansion $(1+x)^N$ follow a fixed ratio determined by the combination formula, allowing us to find the total power $N=n+5$ and the specific position $r$ using the given ratio $5:10:14$ for the terms ${}^N C_{r-1}, {}^N C_r, {}^N C_{r+1}$. By setting up the ratios $\frac{{}^N C_r}{{}^N C_{r-1}} = \frac{N-r+1}{r} = 2$ and $\frac{{}^N C_{r+1}}{{}^N C_r} = \frac{N-r}{r+1} = \frac{14}{10}$, we can establish a system of equations to solve for the unknown variables.

Simplifying these expressions, we obtain $N+1 = 3r$ and $5(N-r) = 7(r+1)$, which leads to $N=11$ and $r=4$ upon substitution. Because the power of the expansion is $N=11$, the largest coefficient occurs at the middle term, specifically at ${}^{11} C_5$ or ${}^{11} C_6$. Evaluating these combinations results in 462, which is the maximum coefficient for this expansion.

To determine the coefficient of $x^4$ in the expansion of $(1 + x + x^2)^{10}$, we represent the expression as a binomial $(1 + (x + x^2))^{10}$ and apply the binomial theorem, $\sum_{k=0}^{n} {}^{n}C_{k} b^k$, where $b = x + x^2$. Each term of the resulting expansion contributes to the final coefficient depending on how it generates the power $x^4$.

The relevant terms that produce $x^4$ occur when $k=2, 3,$ and $4$. For $k=2$, the term is ${}^{10}C_2 (x+x^2)^2 = 45(x^2 + 2x^3 + x^4)$, providing a contribution of 45. For $k=3$, the expansion yields ${}^{10}C_3 x^3(1+x)^3 = 120 x^3(1 + 3x + 3x^2 + x^3)$, where the $3x$ term multiplied by $120x^3$ gives $360x^4$, contributing 360. For $k=4$, the term is ${}^{10}C_4 x^4(1+x)^4 = 210x^4(1 + 4x + \dots)$, which contributes 210. Adding these individual components, $45 + 360 + 210 = 615$, gives the final result.

Pascal's Identity, ${}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r}$, provides the foundation for simplifying this summation by allowing the combination of adjacent terms. Expanding the sequence results in ${}^{50}C_{6} + {}^{49}C_{6} + {}^{48}C_{6} + \dots + {}^{30}C_{6}$. By adding and subtracting ${}^{30}C_{7}$ to the end of this expression, the term ${}^{30}C_{6} + {}^{30}C_{7}$ simplifies into ${}^{31}C_{7}$.

Continuing this telescoping pattern, adding ${}^{31}C_{7}$ to the next term in the series, ${}^{31}C_{6}$, creates ${}^{32}C_{7}$. This chain reaction proceeds systematically up to the final term ${}^{50}C_{6} + {}^{50}C_{7}$, which simplifies to ${}^{51}C_{7}$. Since ${}^{30}C_{7}$ was introduced into the sum to enable these successive simplifications, it must be subtracted at the end, resulting in ${}^{51}C_{7} - {}^{30}C_{7}$.

Determining the term independent of $x$ requires equating the total exponent of $x$ to zero within the general term formula $T_{r+1} = {}^{n}C_{r} a^{n-r} b^r$. For the expression $\left(\frac{3}{2} x^{2} - \frac{1}{3x}\right)^{9}$, the general term is given by $T_{r+1} = {}^{9}C_{r} \left(\frac{3}{2} x^{2}\right)^{9-r} \left(-\frac{1}{3x}\right)^{r}$. Focusing on the exponents of $x$ reveals the condition $x^{18-2r} \cdot x^{-r} = x^{18-3r}$, which implies $18 - 3r = 0$, or $r = 6$. Substituting $r = 6$ provides the value $k = {}^{9}C_{6} \left(\frac{3}{2}\right)^{3} \left(-\frac{1}{3}\right)^{6} = \frac{21}{54}$. Consequently, $18k = \frac{21}{54} \times 18 = 7$.

The summation $S = 1 \cdot ^{25}C_0 + 5 \cdot ^{25}C_1 + 9 \cdot ^{25}C_2 + \dots + 101 \cdot ^{25}C_{25}$ is represented by the general term $(4r + 1) ^{25}C_r$ where $r$ ranges from $0$ to $25$. This sum can be separated into two components: $4 \sum_{r=0}^{25} r \cdot ^{25}C_r$ and \sum_{r=0}^{25} ^{25}C_r. Using the standard binomial identities $\sum_{r=0}^{n} r \cdot ^nC_r = n \cdot 2^{n-1}$ and \sum_{r=0}^{n} ^nC_r = 2^n with $n=25$, the first part evaluates to $4(25 \cdot 2^{24})$ and the second part evaluates to $2^{25}$. Rewriting $4 \cdot 25 \cdot 2^{24}$ as $100 \cdot 2^{24}$, which simplifies to $50 \cdot 2^{25}$, the total sum becomes $50 \cdot 2^{25} + 1 \cdot 2^{25} = 51 \cdot 2^{25}$. Given that the original expression equates this sum to $k \cdot 2^{25}$, the value of $k$ is $51$.

The sum of the coefficients of the even powers in any polynomial $P(x)$ can be calculated by finding the average of the polynomial evaluated at $x = 1$ and $x = -1$, represented by the expression $\frac{P(1) + P(-1)}{2}$. For the product $P(x) = (1 + x + x^2 + \dots + x^{2n})(1 - x + x^2 - x^3 + \dots + x^{2n})$, evaluating the first factor at $x = 1$ gives $1+1+\dots+1$ ($2n+1$ terms), which is $2n+1$, while the second factor evaluates to $1-1+1-1+\dots+1$, which simplifies to 1. Their product $P(1)$ is therefore $2n+1$. Similarly, evaluating the product at $x = -1$, the first factor becomes $1-1+1-1+\dots+1$, resulting in 1, while the second factor becomes $1-(-1)+1-(-1)+\dots+1$, which sums to $2n+1$. Thus, $P(-1)$ is also $2n+1$. Since the sum of the even coefficients is 61, the equation $\frac{(2n+1) + (2n+1)}{2} = 61$ holds, leading to $2n + 1 = 61$, which confirms $n = 30$.

The expression $(1+x+x^2+x^3)^6$ simplifies by factoring the base into $(1+x)(1+x^2)$, allowing the expression to be written as $(1+x)^6(1+x^2)^6$. Expanding these factors using the binomial theorem yields general terms of the form $\binom{6}{r}x^r \cdot \binom{6}{t}x^{2t} = \binom{6}{r}\binom{6}{t}x^{r+2t}$, where the coefficient of $x^4$ is the sum of terms satisfying $r+2t=4$ for non-negative integers $r$ and $t$. The valid integer pairs $(r, t)$ are $(4, 0)$, $(2, 1)$, and $(0, 2)$, which provide individual contributions of $\binom{6}{4}\binom{6}{0} = 15 \times 1 = 15$, $\binom{6}{2}\binom{6}{1} = 15 \times 6 = 90$, and $\binom{6}{0}\binom{6}{2} = 1 \times 15 = 15$. Summing these values, $15 + 90 + 15$, produces a final coefficient of 120.

In the binomial expansion of $\left( \frac{x}{\cos\theta} + \frac{1}{x\sin\theta} \right)^{16}$, we identify the term independent of $x$ by examining the general term $T_{r+1} = {}^{16}C_r \left( \frac{x}{\cos\theta} \right)^{16-r} \left( \frac{1}{x\sin\theta} \right)^r$. The powers of $x$ combine to $x^{16-2r}$, so for the term to be independent of $x$, the exponent $16-2r$ must equal zero, which occurs at $r=8$. This independent term simplifies to ${}^{16}C_8 \frac{1}{\cos^8\theta \sin^8\theta}$, which becomes ${}^{16}C_8 \frac{2^8}{(\sin 2\theta)^8}$ by applying the double-angle identity $\sin 2\theta = 2 \sin\theta \cos\theta$.

To determine the least values of this expression, we must maximize the denominator $(\sin 2\theta)^8$ within the given angular intervals. On the interval $[\frac{\pi}{8}, \frac{\pi}{4}]$, the function $\sin 2\theta$ reaches its maximum value of $1$ at $\theta = \frac{\pi}{4}$, resulting in $l_1 = {}^{16}C_8 \cdot 2^8$. For the interval $[\frac{\pi}{16}, \frac{\pi}{8}]$, the maximum value of $\sin 2\theta$ is $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ at $\theta = \frac{\pi}{8}$, which yields $l_2 = {}^{16}C_8 \frac{2^8}{(1/\sqrt{2})^8} = {}^{16}C_8 \cdot 2^8 \cdot 16$. Dividing $l_2$ by $l_1$ gives a ratio of $16:1$.

The given expression is a finite geometric series containing 126 terms, starting from $49^{0}$ and ending at $49^{125}$. Using the geometric series sum formula, the total value is $\frac{49^{126} - 1}{49 - 1}$, which simplifies to $\frac{49^{126} - 1}{48}$. To identify a factor of the form $49^k + 1$, we can factor the numerator $49^{126} - 1$ by applying the difference of squares identity, $a^2 - b^2 = (a - b)(a + b)$, with $a = 49^{63}$ and $b = 1$. This yields the expression $(49^{63} - 1)(49^{63} + 1)$ for the numerator. Because $49^{63} + 1$ is a factor of the numerator and the remaining term $\frac{49^{63} - 1}{48}$ is an integer, $49^{63} + 1$ must be a factor of the original sum. Therefore, the greatest positive integer $k$ is $63$.

The sum of a series involving binomial coefficients relies on the identities $\sum_{r=0}^{n} \binom{n}{r} = 2^n$ and $\sum_{r=0}^{n} r \binom{n}{r} = n \cdot 2^{n-1}$. For the given series $2 \cdot \binom{20}{0} + 5 \cdot \binom{20}{1} + \dots + 62 \cdot \binom{20}{20}$, the general term is $(2+3r) \binom{20}{r}$ where $r$ ranges from $0$ to $20$, allowing the sum to be separated into $2 \sum_{r=0}^{20} \binom{20}{r} + 3 \sum_{r=0}^{20} r \binom{20}{r}$. Substituting $n=20$ into the standard identities yields $2 \cdot 2^{20} + 3 \cdot 20 \cdot 2^{19}$, which simplifies to $2^{21} + 60 \cdot 2^{19}$ or $4 \cdot 2^{19} + 60 \cdot 2^{19}$. Because $64 \cdot 2^{19}$ is equivalent to $2^6 \cdot 2^{19}$, the final result is $2^{25}$.

The expansion of $(x+10)^{50} + (x-10)^{50}$ follows the binomial theorem, where we isolate specific coefficients by identifying terms associated with powers of $x$. The constant term $a_0$ is found by setting $x=0$, resulting in $10^{50} + (-10)^{50} = 2 \cdot 10^{50}$, whereas the coefficient $a_2$ represents the $x^2$ term, which combines the identical contributions from both binomial expansions to equal $2 \cdot \binom{50}{2} \cdot 10^{48}$.

By setting up the ratio $\frac{a_2}{a_0} = \frac{2 \cdot \binom{50}{2} \cdot 10^{48}}{2 \cdot 10^{50}}$, the expression simplifies to $\frac{\binom{50}{2}}{10^2}$. Given that $\binom{50}{2} = \frac{50 \times 49}{2} = 1225$, the calculation yields $\frac{1225}{100}$, resulting in a final value of 12.25.

To determine the fourth term of the binomial expansion $\left( \frac{2}{x} + x^{\log_{8} x} \right)^6$, one applies the general term formula $T_{r+1} = \binom{n}{r} a^{n-r} b^r$, where $r=3$ for the fourth term. Substituting the given terms into the formula yields $\binom{6}{3} \left( \frac{2}{x} \right)^3 \left( x^{\log_{8} x} \right)^3$, which evaluates to $20 \cdot \frac{8}{x^3} \cdot x^{3 \log_{8} x}$ once the binomial coefficient $\binom{6}{3}$ is calculated.

Setting this expression equal to $20 \times 8^7$ and dividing both sides by $20$ simplifies the equation to $8 \cdot x^{-3} \cdot x^{3 \log_{8} x} = 8^7$, or equivalently $x^{3 \log_{8} x} \cdot x^{-3} = 8^6$. Taking the logarithm with base $8$ on both sides transforms the equation into $3(\log_{8} x)^2 - 3\log_{8} x = 6$. Dividing the entire equation by $3$ results in the quadratic relationship $(\log_{8} x)^2 - \log_{8} x - 2 = 0$, which factors into $(\log_{8} x - 2)(\log_{8} x + 1) = 0$. This provides solutions of $\log_{8} x = 2$ or $\log_{8} x = -1$, leading to $x = 8^2$ or $x = 8^{-1}$. Among the provided choices, $8^2$ is the valid value.

For a binomial expansion of the form $(a + b)^n$ where $n$ is an even number, there is a single middle term located at position $\frac{n}{2} + 1$. In the expression $\left( \frac{x^3}{3} + \frac{3}{x} \right)^8$, the exponent $8$ is even, identifying the middle term as the fifth term in the expansion.

Applying the general term formula $T_{r+1} = {}^nC_r a^{n-r} b^r$ with $n=8$ and $r=4$, the fifth term is calculated as ${}^{8}C_{4} \left(\frac{x^3}{3}\right)^4 \left(\frac{3}{x}\right)^4$. Since ${}^{8}C_{4}$ equals 70 and the algebraic powers simplify to $x^{12}$ divided by $x^4$, the expression reduces to $70x^8$. Equating this to the given value of 5670 yields the equation $70x^8 = 5670$, which simplifies to $x^8 = 81$.

Solving $x^8 = 81$ for real values of $x$ gives $x = \pm \sqrt{3}$ because $(\pm \sqrt{3})^8$ equals $3^4$, which is 81. Since the two real solutions are additive inverses of each other, their sum is 0.

The expression $\left(\frac{1-t^6}{1-t}\right)^3$ expands as the product of the polynomial $(1-t^6)^3$ and the infinite series $(1-t)^{-3}$. Expanding the cubic term yields $1 - 3t^6 + 3t^{12} - t^{18}$, while the negative binomial expansion provides the general term $\binom{k+3-1}{k}t^k$. Since the objective is finding the coefficient of $t^4$, only the product of the constant 1 from the first part and the $t^4$ term from the second part contributes to the final result, as all other terms in the polynomial involve powers of 6 or higher that exceed the required power. The coefficient is thus determined by the calculation $\binom{4+3-1}{4}$, which simplifies to $\binom{6}{2}$ and results in 15.

The binomial theorem reveals that the expansion of $(a + b)^n$ is symmetric, meaning the $k$-th term from the beginning and the $k$-th term from the end are simply the $k$-th terms of $(a + b)^n$ and $(b + a)^n$, respectively. This symmetry ensures that the ratio of these terms relies purely on the powers of the individual components rather than the binomial coefficients, which remain identical for both positions.

For the expression $(2^{1/3} + \frac{1}{2(3^{1/3})})^{10}$, the 5th term from the start is ${}^{10}C_4 (2^{1/3})^6 (\frac{1}{2(3^{1/3})})^4$ and the 5th term from the end is ${}^{10}C_4 (\frac{1}{2(3^{1/3})})^6 (2^{1/3})^4$. Dividing the two expressions causes the ${}^{10}C_4$ coefficients to cancel out, leaving the ratio as:
$\frac{(2^{1/3})^2}{(\frac{1}{2(3^{1/3})})^2} = (2^{1/3} \cdot 2 \cdot 3^{1/3})^2 = (2 \cdot 6^{1/3})^2$
Calculating this value results in $4 \cdot 36^{1/3}$, confirming the ratio is $4(36)^{1/3} : 1$.

The binomial expansion theorem provides the formula $T_{r+1} = {}^n C_r a^{n-r} b^r$ to determine any specific term in an expansion. For the fourth term in this expression, we apply $n=6$ and $r=3$, which yields the coefficient ${}^6 C_3 = 20$.

Setting this term equal to the provided value requires the equation $20 \left( \sqrt{\frac{1}{x^{1+\log_{10} x}}} \right)^3 \left( x^{\frac{1}{12}} \right)^3 = 200$. Dividing both sides by 20 simplifies the equation to $\left( \sqrt{\frac{1}{x^{1+\log_{10} x}}} \cdot x^{\frac{1}{12}} \right)^3 = 10$, which is equivalent to $\sqrt{\frac{1}{x^{1+\log_{10} x}}} \cdot x^{\frac{1}{12}} = 10^{1/3}$.

By rewriting the terms as powers of $x$ and taking the logarithm of both sides to isolate the variable, the equation transforms into a logarithmic form. Solving this resulting equation for $x$ leads to $x=10$, which is the only value that satisfies the condition $x > 1$.

Using the binomial theorem, the $(r+1)$-th term of an expansion $(a+b)^n$ is defined by the formula ${}^nC_r \cdot a^{n-r} \cdot b^r$. For the expansion $(1 + x^{\log_2 x})^5$, the third term corresponds to $r = 2$, which gives ${}^5C_2 \cdot (1)^3 \cdot (x^{\log_2 x})^2$, simplifying to $10 \cdot (x^{\log_2 x})^2$. Setting this equal to 2560 yields $10 \cdot (x^{\log_2 x})^2 = 2560$, or $(x^{\log_2 x})^2 = 256$. Taking the base-2 logarithm of both sides leads to $2 \cdot (\log_2 x)^2 = 8$, which results in $(\log_2 x)^2 = 4$, so $x$ can be 4 or $\frac{1}{4}$.

Finding the term containing a specific power of $x$ begins by applying the binomial theorem to the expression $(\sqrt{x} + \frac{\lambda}{x^2})^{10}$, where the general term is defined as $\binom{10}{r} (\sqrt{x})^{10-r} \left(\frac{\lambda}{x^2}\right)^r$. Simplifying the powers of $x$ within this term, we obtain $\binom{10}{r} x^{\frac{10-r}{2}} \lambda^r x^{-2r}$, which condenses to $\binom{10}{r} \lambda^r x^{\frac{10-5r}{2}}$. Since the entire expansion is multiplied by the leading $x^2$, the final exponent of $x$ becomes $2 + \frac{10-5r}{2}$, which simplifies to $\frac{14-5r}{2}$.

To identify the term where the power of $x$ is 2, we set the exponent $\frac{14-5r}{2} = 2$, which implies $14-5r = 4$, leading to $5r = 10$ and confirming that $r = 2$. Substituting this value of $r$ into the expression for the coefficient, we have $\binom{10}{2} \lambda^2 = 720$. Given that $\binom{10}{2}$ equals 45, the equation becomes $45 \lambda^2 = 720$, which simplifies to $\lambda^2 = 16$, yielding the positive value $\lambda = 4$.

The expansion of $(A+B)^6 + (A-B)^6$ results in twice the sum of terms where the binomial component $B$ is raised to an even power, which simplifies the expression to $2(\binom{6}{0}x^6 + \binom{6}{2}x^4(x^3-1) + \binom{6}{4}x^2(x^3-1)^2 + \binom{6}{6}(x^3-1)^3)$. Substituting the values of the binomial coefficients, $\binom{6}{0}=1$, $\binom{6}{2}=15$, $\binom{6}{4}=15$, and $\binom{6}{6}=1$, transforms the expression into $2(x^6 + 15(x^7-x^4) + 15x^2(x^6-2x^3+1) + (x^9-3x^6+3x^3-1))$. Expanding this polynomial yields $2(x^6 + 15x^7 - 15x^4 + 15x^8 - 30x^5 + 15x^2 + x^9 - 3x^6 + 3x^3 - 1)$. Identifying the terms with even powers of $x$, which are $x^8$, $x^6$, $x^4$, $x^2$, and the constant term $x^0$, their respective coefficients are $15$, $(1-3)=-2$, $-15$, $15$, and $-1$. Summing these coefficients gives $15 - 2 - 15 + 15 - 1 = 12$, and multiplying by the external factor of $2$ results in a total sum of $24$.