Binomial Theorem – JEE Main MathematicsPractice Questions & PYQs

The summation is evaluated using the identity:
$\sum_{r=1}^N r^2 \cdot {}^NC_r = N(N+1)2^{N-2}$
Substituting $N=15$ into this expression yields $15(15+1)2^{15-2} = 15 \cdot 16 \cdot 2^{13} = 15 \cdot 2^4 \cdot 2^{13} = 15 \cdot 2^{17}$. Factorizing the constant $15$ into prime components results in $3^1 \cdot 5^1 \cdot 2^{17}$, which matches the form $2^m \cdot 3^n \cdot 5^k$. Equating the exponents identifies $m=17$, $n=1$, and $k=1$, which gives the sum $m+n+k = 17+1+1 = 19$.

Modular arithmetic allows us to simplify large exponents by examining the remainder of the base when divided by the divisor. Because $64 = 7 \times 9 + 1$, it follows that $64 \equiv 1 \pm od 7$. Since 1 raised to any positive integer power remains 1, raising 64 to the massive exponent given in the expression results in a final remainder of 1 when divided by 7.

The expression $(1919)^{1919}$ can be rewritten as $(1920-1)^{1919}$ to simplify the calculation of its last two digits using the binomial theorem. The expansion is defined as $\sum_{k=0}^{1919} \binom{1919}{k} (1920)^{1919-k} (-1)^k$, which allows for evaluation modulo $100$. Because $1920$ is a multiple of $20$, any term involving $1920^k$ for $k \ge 2$ becomes a multiple of $400$, effectively becoming $0$ modulo $100$.

With these terms removed, only the final two components remain: $\binom{1919}{1918} (1920) - \binom{1919}{1919}$, which simplifies to $1919 \cdot 1920 - 1$. The calculation $3684480 - 1$ leads to $3684479$. Extracting the last two digits, which are $7$ and $9$, the final product is $7 \times 9 = 63$.

The sum of the odd-indexed coefficients $a_1 + a_3 + \dots + a_{19}$ for a polynomial $f(x)$ can be calculated using the identity $\frac{f(1) - f(-1)}{2}$. For the function $f(x) = (1+x+x^2)^{10}$, substituting $x=1$ yields $3^{10} = 59049$, and substituting $x=-1$ yields $(1-1+1)^{10} = 1$. Consequently, the sum of these odd coefficients is $\frac{59049 - 1}{2} = 29524$.

To find the coefficient $a_2$, we utilize the algebraic identity $(1+x+x^2) = \frac{1-x^3}{1-x}$, which implies $f(x) = (1-x^3)^{10}(1-x)^{-10}$. Expanding the components, we have $(1 - 10x^3 + \dots)$ multiplied by the infinite series $\sum_{r=0}^{\infty } \binom{r+9}{9} x^r$. Because we only require the coefficient of $x^2$, the calculation simplifies to finding the $x^2$ term in the second series, which corresponds to $r=2$. This gives $\binom{2+9}{9} = \binom{11}{2} = 55$, meaning $a_2 = 55$.

With these values established, we substitute them into the given equation $29524 - 11(55) = 121 k$. Calculating the product $11 \times 55$ results in $605$, and subtracting this from $29524$ provides $28919$. Solving for $k$ by dividing $28919$ by $121$ results in $k = 239$.

The binomial expansion of $(x+\sqrt{x^3-1})^5 + (x-\sqrt{x^3-1})^5$ relies on the property that when raising the sum and difference of the same two terms to an odd power, the terms involving odd powers of the radical term cancel out. The expression simplifies to twice the sum of the even-indexed terms in the binomial expansion, written as $2[{}^5 C_0 x^5 + {}^5 C_2 x^3 (\sqrt{x^3-1})^2 + {}^5 C_4 x (\sqrt{x^3-1})^4]$. Expanding this yields $2[x^5 + 10x^3(x^3-1) + 5x(x^3-1)^2]$, which simplifies further to $2[x^5 + 10x^6 - 10x^3 + 5x(x^6 - 2x^3 + 1)]$. Distributing the factor of 2 results in the polynomial $10x^7 + 20x^6 + 2x^5 - 20x^4 - 20x^3 + 10x$. Identifying the coefficients based on the powers of $x$, we find $\alpha = 10$, $\beta = 2$, $\gamma = -20$, and $\delta = 10$. Substituting these values into the given conditions creates the following system of linear equations:
$10u + 2v = 18$
$-20u + 10v = 20$
Dividing the second equation by 2 gives $-10u + 5v = 10$. Adding this simplified second equation to the first equation results in $7v = 28$, which determines that $v = 4$. Substituting this value back into the first equation yields $10u + 2(4) = 18$, simplifying to $10u = 10$, so $u = 1$. The sum $u + v$ is therefore $1 + 4 = 5$.

When three consecutive coefficients of a binomial expansion form an arithmetic progression, the middle coefficient must equal the arithmetic mean of the two adjacent terms. For the expansion of $(1+x)^{n+4}$, the coefficients of the 5th, 6th, and 7th terms are ${}^{n+4}C_4$, ${}^{n+4}C_5$, and ${}^{n+4}C_6$, establishing the relationship ${}^{n+4}C_4 + {}^{n+4}C_6 = 2 \times {}^{n+4}C_5$. Solving this algebraic condition leads to the quadratic equation $n^2 - 13n + 30 = 0$, resulting in $n=3$ or $n=10$.

Given the constraint $n \neq 10$, we use $n=3$, which simplifies the expression to $(1+x)^7$. The greatest binomial coefficient in an expansion occurs at the middle term, so for an exponent of 7, this corresponds to ${}^{7}C_3$ or ${}^{7}C_4$. Calculating this central value yields $\frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.

To simplify the expression, we focus on the general term $\frac{r^2 \left({}^{n}C_r\right)^2}{{}^{n}C_{r-1}}$ where $n=30$. By rewriting the ratio $\frac{{}^nC_r}{{}^nC_{r-1}}$ as $\frac{n-r+1}{r}$, the term simplifies to $r^2 \cdot {}^nC_r \cdot \frac{n-r+1}{r}$, which reduces to $r(n-r+1) \cdot {}^nC_r$. Using the property $r \cdot {}^nC_r = n \cdot {}^{n-1}C_{r-1}$, we substitute this to obtain $n(n-r+1) \cdot {}^{n-1}C_{r-1}$. Setting $n=30$, the original summation becomes $\sum_{r=1}^{30} 30(31-r) {}^{29}C_{r-1}$.

By changing the variable to $k = r-1$, where $k$ ranges from $0$ to $29$, the expression transforms into $30 \sum_{k=0}^{29} (30-k) {}^{29}C_k$. This can be separated into two distinct sums: $30 \left( 30 \sum_{k=0}^{29} {}^{29}C_k - \sum_{k=0}^{29} k {}^{29}C_k \right)$. Applying the fundamental binomial identities $\sum_{k=0}^{m} {}^{m}C_k = 2^m$ and $\sum_{k=0}^{m} k \cdot {}^{m}C_k = m \cdot 2^{m-1}$ with $m=29$, we compute the expression as $30 \left( 30 \cdot 2^{29} - 29 \cdot 2^{28} \right)$. Factoring out $2^{28}$, this becomes $30 \cdot 2^{28} (60 - 29)$, which simplifies to $30 \cdot 31 \cdot 2^{28} = 930 \cdot 2^{28}$. Writing this as $465 \cdot 2 \cdot 2^{28}$ gives $465 \cdot 2^{29}$, confirming that $\alpha$ is 465.

The general term in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^n C_r a^{n-r} b^r$. The $15^{th}$ term from the beginning is $T_{15} = {}^n C_{14} (2^{\frac{1}{3}})^{n-14} (3^{-\frac{1}{3}})^{14}$, while the $15^{th}$ term from the end corresponds to the $15^{th}$ term of the expansion $(\frac{1}{\sqrt[3]{3}} + \sqrt[3]{2})^n$, which is $T'_{15} = {}^n C_{14} (3^{-\frac{1}{3}})^{n-14} (2^{\frac{1}{3}})^{14}$.

Dividing $T_{15}$ by $T'_{15}$ yields:
$\frac{T_{15}}{T'_{15}} = \frac{{}^n C_{14} (2^{\frac{n-14}{3}}) (3^{-\frac{14}{3}})}{{}^n C_{14} (3^{-\frac{n-14}{3}}) (2^{\frac{14}{3}})} = \frac{2^{\frac{n-14}{3}} \cdot 2^{-\frac{14}{3}}}{3^{-\frac{n-14}{3}} \cdot 3^{\frac{14}{3}}} = \frac{2^{\frac{n-28}{3}}}{3^{-\frac{n-28}{3}}} = 6^{\frac{n-28}{3}}$

Equating this ratio to the given value $\frac{1}{6}$, we have $6^{\frac{n-28}{3}} = 6^{-1}$. Solving the exponent equation $\frac{n-28}{3} = -1$ gives $n-28 = -3$, which results in $n=25$. Calculating the combination value for $n=25$ gives:
${}^{25} C_3 = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 25 \times 4 \times 23 = 2300$

To find the term independent of $x$, begin by simplifying the complex expression inside the parentheses. Recognizing the algebraic identity for the sum of cubes, $a^3+b^3 = (a+b)(a^2-ab+b^2)$, the first fraction $\frac{x+1}{x^{2/3}-x^{1/3}+1}$ simplifies to $\frac{(x^{1/3}+1)(x^{2/3}-x^{1/3}+1)}{x^{2/3}-x^{1/3}+1}$, which reduces to $x^{1/3}+1$. Similarly, the second fraction $\frac{x-1}{x-x^{1/2}}$ can be rewritten as $\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)}$, which simplifies to $\frac{\sqrt{x}+1}{\sqrt{x}}$, or $1+x^{-1/2}$. Subtracting these terms results in $(x^{1/3}+1) - (1+x^{-1/2}) = x^{1/3}-x^{-1/2}$.

With the expression now reduced to $(x^{1/3}-x^{-1/2})^{10}$, the general term is given by:

$T_{r+1} = \binom{10}{r} (x^{1/3})^{10-r} (-x^{-1/2})^r = \binom{10}{r} (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}}$

For the term to be independent of $x$, the net exponent of $x$ must be zero. Setting $\frac{10-r}{3} - \frac{r}{2} = 0$ leads to the equation $20 - 2r - 3r = 0$, which simplifies to $5r = 20$, yielding $r=4$. Substituting $r=4$ into the binomial coefficient gives $\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}$, which equals 210.

In the binomial expansion of $(2+\sqrt{3})^8$, any given term follows the general form $T_{r+1} = {}^{8}C_{r} (2)^{8-r} (\sqrt{3})^r$. A term yields a rational number if and only if the exponent of the irrational part, $\sqrt{3}$, is an even number, ensuring that powers like $(\sqrt{3})^2$, $(\sqrt{3})^4$, $(\sqrt{3})^6$, and $(\sqrt{3})^8$ simplify to integers. Consequently, the rational terms in this series occur specifically when $r$ takes the values $0, 2, 4, 6,$ and $8$.

Evaluating each of these individually, for $r=0$, the term is ${}^{8}C_{0} \times 2^8 = 256$. For $r=2$, the calculation is ${}^{8}C_{2} \times 2^6 \times (\sqrt{3})^2 = 28 \times 64 \times 3 = 5376$. Moving to $r=4$, the term is ${}^{8}C_{4} \times 2^4 \times (\sqrt{3})^4 = 70 \times 16 \times 9 = 10080$. When $r=6$, the term is ${}^{8}C_{6} \times 2^2 \times (\sqrt{3})^6 = 28 \times 4 \times 27 = 3024$. Finally, for $r=8$, the term is ${}^{8}C_{8} \times (\sqrt{3})^8 = 1 \times 81 = 81$. Summing these values together, $256 + 5376 + 10080 + 3024 + 81$, results in a total of $18817$.

The general term $T_{k+1}$ in the expansion of $(\sqrt[3]{7}+\sqrt[12]{11})^n$ is given by $\binom{n}{k} 7^{\frac{n-k}{3}} 11^{\frac{k}{12}}$, where integral terms require both exponents to be integers. This condition implies that $k$ must be a multiple of $12$, which ensures $k$ is also a multiple of $3$, and subsequently requires $n$ to be a multiple of $3$. With the sequence of $k$ values being $0, 12, 24, \dots, 12m$, having 183 terms means $m+1 = 183$, so the maximum $k$ is $12 \times 182 = 2184$. Since $k$ must be less than or equal to $n$ and $n$ must be divisible by $3$, the least value for $n$ is $2184$.

The coefficient of the $(r+1)^{\text{th}}$ term in the binomial expansion of $(1+x)^m$ is defined by the combination ${}^mC_r$. For the expansion of $(1+x)^{2n-1}$, the coefficient $A$ of the 30th term is ${}^{2n-1}C_{29}$ and the coefficient $B$ of the 12th term is ${}^{2n-1}C_{11}$. Substituting these expressions into the given condition $2A=5B$ yields $2 \cdot {}^{2n-1}C_{29} = 5 \cdot {}^{2n-1}C_{11}$, which can be rearranged to the ratio $\frac{{}^{2n-1}C_{29}}{{}^{2n-1}C_{11}} = \frac{5}{2}$. Using the binomial property $\frac{{}^mC_r}{{}^mC_{r-1}} = \frac{m-r+1}{r}$ and substituting $n=21$ results in a total index of $2n-1=41$, where the ratio $\frac{{}^{41}C_{29}}{{}^{41}C_{11}}$ simplifies to the required value of 2.5, confirming that $n=21$.

The condition for the coefficients of three consecutive terms in a binomial expansion, denoted as $\binom{12}{r-1}, \binom{12}{r}, \binom{12}{r+1}$, to form a geometric progression requires that $(\binom{12}{r})^2 = \binom{12}{r-1} \binom{12}{r+1}$. Because no set of three consecutive binomial coefficients can exist in a geometric progression, there are no possible values for $r$, which means $p=0$.

For the expansion $(\sqrt[4]{3} + \sqrt[3]{4})^{12}$, any term is defined by the general expression $T_{k+1} = \binom{12}{k} (3^{\frac{12-k}{4}}) (4^{\frac{k}{3}})$. A term is rational only if the exponent of each base is an integer, which requires $12-k$ to be divisible by $4$ and $k$ to be divisible by $3$. Evaluating $k$ within the valid range of $0$ to $12$, the conditions are satisfied only when $k=0$ or $k=12$. Substituting these values yields $T_1 = \binom{12}{0} (3^3)(4^0) = 27$ and $T_{13} = \binom{12}{12} (3^0)(4^4) = 256$. Summing these rational terms provides the value $q = 27 + 256 = 283$, resulting in the final sum $p+q = 283$.

To evaluate the summation, decompose the expression by linearity into the sum of two components: $\sum_{r=1}^9 \frac{r}{2^r} \cdot {}^9 C_r$ and $\sum_{r=1}^9 \frac{3}{2^r} \cdot {}^9 C_r$. For the first component, apply the identity $r \cdot {}^n C_r = n \cdot {}^{n-1} C_{r-1}$, which allows the expression to be written as $9 \sum_{r=1}^9 {}^8 C_{r-1} \left(\frac{1}{2}\right)^r$. Factoring out $\frac{1}{2}$ transforms this into $\frac{9}{2} \sum_{r=1}^9 {}^8 C_{r-1} \left(\frac{1}{2}\right)^{r-1}$, which simplifies to the binomial form $\frac{9}{2} (1+\frac{1}{2})^8 = \frac{9}{2} \cdot \frac{3^8}{2^8} = \frac{3^{10}}{2^9}$.

For the second component, factor out the constant 3 to obtain $3 \sum_{r=1}^9 {}^9 C_r \left(\frac{1}{2}\right)^r$. This series represents the binomial expansion of $(1+\frac{1}{2})^9$ missing the $r=0$ term, giving $3 \left[ (1+\frac{1}{2})^9 - {}^9 C_0 \left(\frac{1}{2}\right)^0 \right]$ = 3 \left[ \left(\frac{3}{2}\right)^9 - 1 \right]$ = 3 \left(\frac{3}{2}\right)^9 - 3$. Combining both components results in$\frac{3^{10}}{2^9} + 3 \left(\frac{3}{2}\right)^9 - 3$. Recognizing that$\frac{3^{10}}{2^9} = 3 \left(\frac{3^9}{2^9}\right) = 3 \left(\frac{3}{2}\right)^9$, the total expression becomes$3 \left(\frac{3}{2}\right)^9 + 3 \left(\frac{3}{2}\right)^9 - 3$, which simplifies to$6 \left(\frac{3}{2}\right)^9 - 3$. Comparing this to$\alpha \left(\frac{3}{2}\right)^9 - \beta$, we find$\alpha = 6$and$\beta = 3$, thus$(\alpha + \beta)^2 = (6+3)^2 = 81$.

Fermat's Little Theorem provides an elegant starting point, asserting that for a prime $p$ and an integer $a$ not divisible by $p$, the congruence $a^{p-1} \equiv 1 \pm od{p}$ must hold. With $p=23$ and $a=7$, we obtain $7^{22} \equiv 1 \pm od{23}$. The large exponent 103 is simplified by dividing it by 22, which yields a quotient of 4 and a remainder of 15, allowing the expression to be written as $7^{103} = (7^{22})^4 \cdot 7^{15} \equiv 1^4 \cdot 7^{15} \equiv 7^{15} \pm od{23}$.

Proceeding with modular exponentiation, we observe that $7^3 = 343$, and dividing 343 by 23 leaves a remainder of -2. Expressing $7^{15}$ as $(7^3)^5$ transforms the calculation into $(-2)^5 \pm od{23}$. Because $(-2)^5 = -32$ and $-32 \equiv -32 + (2 \times 23) = 14 \pm od{23}$, the final remainder is 14.

The general term in the binomial expansion of $(a+b)^n$ is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. Applying this to the expression $\left(5^{\frac{1}{2}}+7^{\frac{1}{8}}\right)^{1016}$, we define the general term as:

$T_{r+1} = \binom{1016}{r} 5^{\frac{1016-r}{2}} 7^{\frac{r}{8}}$

For this term to be an integer, both exponents $\frac{1016-r}{2}$ and $\frac{r}{8}$ must be non-negative integers. Because any multiple of 8 is necessarily an even number, requiring $r$ to be a multiple of 8 automatically ensures that the exponent of 5 also results in an integer. Calculating the values of $r=8k$ within the range $0 \le r \le 1016$ requires finding all integers $k$ such that $0 \le k \le 127$, resulting in a total of 128 integral terms.

The multinomial expansion of $\left(1+2^{\frac{1}{3}}+3^{\frac{1}{2}}\right)^6$ consists of general terms of the form $\frac{6!}{\alpha!\beta!\gamma!} (1)^\alpha (2^{\frac{1}{3}})^\beta (3^{\frac{1}{2}})^\gamma$, where $\alpha, \beta, \gamma$ are non-negative integers summing to 6. For a term to be rational, the fractional powers must reduce to integers, requiring $\beta$ to be a multiple of 3 and $\gamma$ to be a multiple of 2. By systematically evaluating the valid combinations of $\beta \in \{0, 3, 6\}$ and $\gamma \in \{0, 2, 4, 6\}$ such that $\alpha+\beta+\gamma=6$, we can identify every rational component.

When $\gamma=0$, the cases for $\beta=0, 3, 6$ result in terms of 1, 40, and 4. When $\gamma=2$, setting $\beta=0$ and $\beta=3$ yields terms of 45 and 360. For $\gamma=4$, only $\beta=0$ is possible, giving a term of 135, and for $\gamma=6$, setting $\beta=0$ gives a term of 27. Summing all these rational values, the total is $1 + 40 + 4 + 45 + 360 + 135 + 27 = 612$.

This problem relies on manipulating the binomial expansion $(1+x)^{20} = \sum_{k=0}^{20} {}^{20}C_k x^k$ to produce the coefficients $(k-2)(k-3)$ associated with ${}^{20}C_k$. By dividing the entire binomial expression by $x^2$, we obtain a series where the general term becomes ${}^{20}C_k x^{k-2}$. Differentiating this function twice with respect to $x$ introduces the factor $(k-2)(k-3)$ to the coefficients, creating terms that match the structure of the series. Specifically, the second derivative produces a series where the term for $k=4$ is $2 \cdot 1 \cdot {}^{20}C_4$, the term for $k=5$ involves $3 \cdot 2 \cdot {}^{20}C_5 x$, and so on. Evaluating this differentiated expression at $x = -1$ naturally introduces the alternating signs required by the series, as the powers of $x$ (such as $x^0, x^1, x^2$) interact with the negative base. Simplifying the resulting equation forces the alternating sum to equal 34.

Integrating the binomial expansion of $(1+x)^{11} = \sum_{k=0}^{11} {}^{11}C_k x^k$ provides a direct path to the required series. By evaluating the definite integral $\int_{0}^{1} (1+x)^{11} \, dx = \left[ \frac{(1+x)^{12}}{12} \right]_0^1 = \frac{2^{12}-1}{12}$, and simultaneously integrating the expansion term-by-term as $\sum_{k=0}^{11} {}^{11}C_k \int_{0}^{1} x^k \, dx = \sum_{k=0}^{11} \frac{{}^{11}C_k}{k+1}$, one obtains the sum of all terms from $k=0$ to $11$. This establishes that $\sum_{k=0}^{11} \frac{{}^{11}C_k}{k+1} = \frac{2^{12}-1}{12}$.

To isolate only the odd-indexed terms needed for the target sum, consider the integral over the interval $[-1, 0]$, where $\int_{-1}^{0} (1+x)^{11} \, dx = \left[ \frac{(1+x)^{12}}{12} \right]_{-1}^0 = \frac{1}{12}$. Expanding the integral term-by-term within this interval gives $\sum_{k=0}^{11} {}^{11}C_k \left[ \frac{x^{k+1}}{k+1} \right]_{-1}^0 = \sum_{k=0}^{11} \frac{{}^{11}C_k (-1)^k}{k+1} = \frac{1}{12}$. Subtracting this second equation from the first causes the even-indexed terms to cancel while the odd-indexed terms add together: $\frac{2^{12}-1}{12} - \frac{1}{12} = 2 \sum_{r=0}^{5} \frac{{}^{11}C_{2r+1}}{2r+2}$. Simplifying the expression leads to $\frac{2^{12}-2}{12} = 2 \sum_{r=0}^{5} \frac{{}^{11}C_{2r+1}}{2r+2}$, which yields $\sum_{r=0}^{5} \frac{{}^{11}C_{2r+1}}{2r+2} = \frac{2^{11}-1}{12} = \frac{2047}{12}$. With $m = 2047$ and $n = 12$ being coprime, the value of $m-n$ is $2047 - 12 = 2035$.

The sum of the coefficients in the binomial expansion of $(x+y)^m$ is $2^m$, and since there are $m+1$ coefficients, the arithmetic mean is $\frac{2^m}{m+1} = 16$. Substituting $m=2n-3$ gives $\frac{2^{2n-3}}{2n-2} = 16$, which simplifies to $2^{2n-3} = 16 \times 2(n-1) = 2^5(n-1)$, or $2^{2n-8} = n-1$. Testing integer values shows that $n=5$ is the solution because $2^{10-8} = 4 = 5-1$. For $n=5$, the coordinates of point $P$ are $(2(5)-1, 5^2-4(5)) = (9, 5)$. The perpendicular distance $d$ from a point $(x_1, y_1)$ to the line $Ax+By+C=0$ is given by $d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$. Substituting $x_1=9$, $y_1=5$, and the line equation $x+y-8=0$, the distance is $d = \frac{|9+5-8|}{\sqrt{1^2+1^2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$.