Binomial Theorem – JEE Main MathematicsPractice Questions & PYQs

Recognizing the structure of the summation reveals it to be the expansion of the coefficient of $x^r$ in the product of two identical binomials, $(1+x)^{20}$ multiplied by $(1+x)^{20}$. This product simplifies directly to the single binomial expansion $(1+x)^{40}$, making the entire summation equivalent to the binomial coefficient $\binom{40}{r}$. Since the binomial coefficient $\binom{n}{r}$ achieves its maximum value when $r$ is exactly at the midpoint of $n$, calculating $r = 40/2$ confirms that the sum is maximized when $r = 20$.

The expression $(1+x)(1-x)^{10} (1+x+x^{2})^{9}$ simplifies by recognizing the identity $(1-x)(1+x+x^{2}) = (1-x^{3})$, which allows the product to be rewritten as $(1+x)(1-x)(1-x)^{9} (1+x+x^{2})^{9} = (1-x^{2})(1-x^{3})^{9}$. Expanding $(1-x^{3})^{9}$ using the binomial theorem gives $\sum_{k=0}^{9} \binom{9}{k} (-1)^{k} x^{3k}$, making the full product $(1-x^{2}) \sum_{k=0}^{9} \binom{9}{k} (-1)^{k} x^{3k}$. To obtain the coefficient of $x^{18}$, we look for terms where $3k=18$, which yields $k=6$ and results in $\binom{9}{6} = 84$, while the term involving $-x^{2}$ produces no $x^{18}$ term because no integer $k$ satisfies $3k+2=18$. Therefore, the total coefficient is $84$.

The summation $\sum_{r=1}^{n} r^2 \cdot {}^{n}C_r$ is determined by the general combinatorial identity $n(n+1)2^{n-2}$. Substituting $n=20$ into this expression yields $20(20+1)2^{20-2}$. This results in the product $20 \times 21 \times 2^{18}$, which simplifies to $420(2)^{18}$. Matching this final value to $A(2^{\beta})$ confirms $A=420$ and $\beta=18$, providing the ordered pair $(420, 18)$.

The core principle underlying this problem is the application of the binomial theorem to sum the components of a geometric series. By representing the individual geometric sums $S_{k-1}$ as $\frac{1-q^k}{1-q}$, the entire expression can be treated as a combination of two binomial expansions, which allows for a straightforward algebraic simplification.

We begin by rewriting the given sum as $\sum_{k=1}^{101} {}^{101}C_k \left( \frac{1-q^k}{1-q} \right)$. Factoring out the term $\frac{1}{1-q}$ allows us to distribute the summation, resulting in $\frac{1}{1-q} \left[ \sum_{k=1}^{101} {}^{101}C_k - \sum_{k=1}^{101} {}^{101}C_k q^k \right]$. The first summation, $\sum_{k=1}^{101} {}^{101}C_k$, is simply the expansion of $(1+1)^{101}$ excluding the first term where $k=0$, yielding $2^{101}-1$. Similarly, the second summation, $\sum_{k=1}^{101} {}^{101}C_k q^k$, is the expansion of $(1+q)^{101}$ excluding the first term, giving $(1+q)^{101}-1$. Subtracting these two results provides the expression $\frac{2^{101}-(1+q)^{101}}{1-q}$ for the entire left side of the equation.

Turning to the right side, the term $\alpha T_{100}$ represents the product of $\alpha$ and a geometric series $\sum_{j=0}^{100} \left( \frac{q+1}{2} \right)^j$. Applying the sum formula for a finite geometric progression, this becomes $\alpha \left( \frac{1-(\frac{q+1}{2})^{101}}{1-(\frac{q+1}{2})} \right)$. Simplifying the denominator $1-\frac{q+1}{2}$ to $\frac{1-q}{2}$, the expression becomes $\alpha \left( \frac{1-\frac{(1+q)^{101}}{2^{101}}}{\frac{1-q}{2}} \right)$, which simplifies further to $\alpha \left( \frac{2}{1-q} \cdot \frac{2^{101}-(1+q)^{101}}{2^{101}} \right)$. Reducing this gives $\alpha \frac{2^{101}-(1+q)^{101}}{(1-q)2^{100}}$. By comparing this result to the previously simplified left side, it is clear that $\alpha$ must equal $2^{100}$ for the two sides to be identical.

Expanding $(1-3x)^{15}$ with the binomial theorem yields $1 - 45x + 945x^2 - 12285x^3 + \dots$. Multiplying this expression by $(1+ax+bx^2)$ allows for the identification of the $x^2$ and $x^3$ coefficients by grouping like terms. The coefficient of $x^2$ is $945 - 45a + b$, which must equal zero, leading to the linear equation $45a - b = 945$. Similarly, the coefficient of $x^3$ is given by $-12285 + 945a - 45b$, and setting this to zero simplifies to $21a - b = 273$. Subtracting these equations yields $24a = 672$, which determines $a = 28$. Substituting this value into the equation $21a - b = 273$ results in $588 - b = 273$, giving $b = 315$. The resulting ordered pair is $(a, b) = (28, 315)$.

The binomial expansion of $(x+1)^n$ relies on the properties of combinations, where consecutive coefficients are expressed as ${}^nC_{r-1}$, ${}^nC_r$, and ${}^nC_{r+1}$. Given that the ratio of these consecutive terms is $2 : 15 : 70$, we can determine the specific values for $n$ and $r$ by solving the ratio of successive combinations, which leads to $n=16$ and $r=2$. Substituting these values into the sequence, we find the three specific coefficients to be ${}^{16}C_1$, ${}^{16}C_2$, and ${}^{16}C_3$, which evaluate to 16, 120, and 560 respectively. Taking the average of these three values results in the expression $\frac{16+120+560}{3}$, which simplifies to 232.

The constant term in this product is found by distributing the terms $\left( \frac{1}{60} - \frac{x^{8}}{81} \right)$ across the binomial expansion of $\left( 2 x^{2} - \frac{3}{x^{2}} \right)^{6}$ and identifying which products result in a net power of $x^0$. Multiplying $\frac{1}{60}$ by the term within the binomial expansion where the powers of $x$ cancel perfectly yields a value of $-72$.

Additionally, multiplying $-\frac{x^8}{81}$ by the portion of the binomial expansion that contains $x^{-8}$ eliminates the variable, producing a value of $36$. Adding these two resulting constants together determines the final value of $-36$.

Pascal's identity allows us to simplify the denominator ${}^{20}C_i + {}^{20}C_{i-1}$ directly to ${}^{21}C_i$. This transformation turns the fraction inside the summation into the ratio $\frac{{}^{20}C_{i-1}}{{}^{21}C_i}$, which, by the properties of binomial coefficients, reduces to $\frac{i}{21}$. Substituting this result into the original problem changes the equation to the following:

$\sum_{i=1}^{20} \left( \frac{i}{21} \right)^3 = \frac{k}{21}$

Factoring out the constant $\frac{1}{21^3}$ from the summation leaves us with the sum of the first 20 cubes, $\sum_{i=1}^{20} i^3$. We calculate this using the standard formula $\left[ \frac{n(n+1)}{2} \right]^2$. Applying this for $n=20$, we find $\left[ \frac{20 \times 21}{2} \right]^2 = (10 \times 21)^2$, which expands to $100 \times 21^2$.

To finish, we divide this resulting sum by the $21^3$ factor we pulled out earlier:

$\frac{100 \times 21^2}{21^3} = \frac{100}{21}$

Comparing this simplified result to the original expression $\frac{k}{21}$, we see that $k$ must equal 100.

Determining the fractional part of $\frac{2^{403}}{15}$ requires expressing the numerator in a form related to the denominator. Since 2 raised to the power of 4 is 16, which is one greater than 15, we rewrite $2^{403}$ as $8 \cdot (2^4)^{100}$, or $8 \cdot 16^{100}$. By writing 16 as $15+1$, the expression becomes $8 \cdot (15+1)^{100}$. Expanding $(15+1)^{100}$ using the binomial theorem results in a sum where every term contains a factor of 15 except for the final constant, represented as $(15\lambda + 1)$ for some integer $\lambda$. Distributing the 8 gives $8 \cdot 15\lambda + 8$, and dividing this entire quantity by 15 yields $8\lambda + \frac{8}{15}$. As $8\lambda$ is an integer, the fractional part is $\frac{8}{15}$, which directly implies $k = 8$.

The general term in the expansion of $\left(x^2 + \frac{1}{x^3}\right)^n$ is $T_{r+1} = n_{C_r} x^{2n-5r}$. Since we are looking for the coefficient of $x$, the exponent must be one, leading to $2n - 5r = 1$. The problem specifies the coefficient as $n_{C_{23}}$, and given the symmetry of binomial coefficients, $n_{C_r} = n_{C_{23}}$ holds when $r = 23$ or $r = n - 23$. Substituting $r = 23$ into the exponent equation gives $2n - 115 = 1$, resulting in $n = 58$. Conversely, setting $r = n - 23$ leads to $2n - 5(n - 23) = 1$, which simplifies to $3n = 114$ and provides $n = 38$. Therefore, the smallest natural number $n$ that satisfies the condition is 38.

The binomial expansion of an expression with radical bases contains a mix of rational and irrational terms, and identifying the rational ones is the key to finding the irrational count. For the expression $\left(7^{\frac{1}{5}}-3^{\frac{1}{10}}\right)^{60}$, the general term is represented by the formula:
$T_{r+1} = {}^{60}C_r \cdot (7)^{\frac{60-r}{5}} \cdot (3)^{\frac{r}{10}} \cdot (-1)^r$
For a term to be rational, the exponents of 7 and 3 must result in integers. This requires $\frac{60-r}{5}$ and $\frac{r}{10}$ to be integers, which implies $r$ must be a multiple of 10 to satisfy both conditions simultaneously within the range $0 \leq r \leq 60$. Testing these, $r$ can be $0, 10, 20, 30, 40, 50, 60$, which yields 7 rational terms in total. Since the expansion of a binomial to the power of 60 contains 61 terms, subtracting these 7 rational terms leaves 54 irrational terms.

The general term for the product of these expansions is given by $\binom{2}{a} \binom{3}{b} \binom{4}{c} x^{a+2b+3c}$, where the indices satisfy the constraints $0 \le a \le 2$, $0 \le b \le 3$, and $0 \le c \le 4$. To find the coefficient of $x^{10}$, we solve the equation $a+2b+3c=10$ for valid integer values of $a$, $b$, and $c$.

For $c=3$, the equation $a+2b=1$ is satisfied only by the set $(a=1, b=0)$, which provides a coefficient of $\binom{2}{1} \binom{3}{0} \binom{4}{3} = 2 \times 1 \times 4 = 8$. For $c=2$, the equation $a+2b=4$ is satisfied by two sets: $(a=2, b=1)$ and $(a=0, b=2)$. These contribute $\binom{2}{2} \binom{3}{1} \binom{4}{2} = 1 \times 3 \times 6 = 18$ and $\binom{2}{0} \binom{3}{2} \binom{4}{2} = 1 \times 3 \times 6 = 18$ respectively. For $c=1$, the equation $a+2b=7$ is satisfied only by the set $(a=1, b=3)$, which provides a coefficient of $\binom{2}{1} \binom{3}{3} \binom{4}{1} = 2 \times 1 \times 4 = 8$. Summing these individual coefficients, we get $8 + 18 + 18 + 8 = 52$.

Expanding $(x+\sqrt{x^3-1})^5+(x-\sqrt{x^3-1})^5$ is most efficiently handled by utilizing the property that adding conjugate binomial expansions eliminates terms with odd powers of the radical, effectively doubling the even-indexed terms in the expansion. Applying the binomial theorem to this expression yields $2(x^5 + 10x^3(x^3-1) + 5x(x^3-1)^2)$. Distributing the terms results in $2x^5 + 20x^6 - 20x^3 + 10x(x^6 - 2x^3 + 1)$, which further simplifies to $2x^5 + 20x^6 - 20x^3 + 10x^7 - 20x^4 + 10x$. Collecting the odd-degree terms $10x^7$, $2x^5$, $-20x^3$, and $10x$ allows us to sum their coefficients as $10 + 2 - 20 + 10$, which results in 2.

The coefficient of $x^2$ in the product $(2-x^2)(P(x)+Q(x))$ is determined by calculating $2 \times (\text{coefficient of } x^2 \text{ in } S(x)) - 1 \times (\text{coefficient of } x^0 \text{ in } S(x))$, where $S(x) = (1+2x+3x^2)^6+(1-4x^2)^6$. Using the binomial theorem, the expansion of $(1+(2x+3x^2))^6$ yields a constant term of $1$ and an $x^2$ coefficient of $6(3) + \binom{6}{2}(2)^2 = 18 + 60 = 78$. The expansion of $(1-4x^2)^6$ yields a constant term of $1$ and an $x^2$ coefficient of $6(-4) = -24$. Summing these components, the total constant term of $S(x)$ is $2$ and the total coefficient of $x^2$ is $78 - 24 = 54$. Applying these values to the original expression, the coefficient of $x^2$ is $2(54) - 1(2) = 108 - 2 = 106$.

Express the base 27 as $28 - 1$, which is equivalent to -1 in modulo 7 since 28 is a multiple of 7. Raising this to the power of 999 creates the expression $(-1)^{999}$, and because the exponent is odd, the value simplifies to -1. Adding the divisor 7 to this negative result yields the positive remainder 6.

The expression can be decomposed into two separate series, $\sum_{r=1}^{10} {}^{21}C_r$ and $\sum_{r=1}^{10} {}^{10}C_r$, by distributing the summation. The second series represents the sum of binomial coefficients for $n=10$ excluding the ${}^{10}C_0$ term, which simplifies to $2^{10}-1$.

For the first series, applying the symmetry property ${}^{n}C_r = {}^{n}C_{n-r}$ to the identity $\sum_{r=0}^{21} {}^{21}C_r = 2^{21}$ reveals that the sum of the first ten coefficients is half of the total sum after excluding the edge terms ${}^{21}C_0$ and ${}^{21}C_{21}$, leading to $\sum_{r=1}^{10} {}^{21}C_r = \frac{1}{2}(2^{21}-2) = 2^{20}-1$. Finally, subtracting the second series from the first yields $(2^{20}-1)-(2^{10}-1)$, which simplifies to $2^{20}-2^{10}$.

The expression $\frac{x+1}{x^{2/3}-x^{1/3}+1}$ simplifies using the algebraic identity $a^3+b^3 = (a+b)(a^2-ab+b^2)$ with $a=x^{1/3}$ and $b=1$, resulting in $x^{1/3}+1$. Similarly, the term $\frac{x-1}{x-x^{1/2}}$ simplifies by factoring $x-1 = (x^{1/2}-1)(x^{1/2}+1)$ and $x-x^{1/2} = x^{1/2}(x^{1/2}-1)$, which yields $\frac{x^{1/2}+1}{x^{1/2}} = 1+x^{-1/2}$. Subtracting these simplified terms produces the binomial expression $(x^{1/3} - x^{-1/2})^{10}$.

The general term in the expansion is $T_{r+1} = \binom{10}{r} (x^{1/3})^{10-r} (-x^{-1/2})^r$, which simplifies to $\binom{10}{r} (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}}$. Equating the exponent of $x$ to $-5$ gives $\frac{10-r}{3} - \frac{r}{2} = -5$, leading to $20 - 2r - 3r = -30$. Solving this equation results in $5r = 50$, so $r=10$. Substituting $r=10$ into the coefficient part of the general term yields $\binom{10}{10}(-1)^{10} = 1$.

To determine coefficients in binomial expansions, we utilize the general term formula $T_{r+1} = {}^{18}C_r a^{18-r} b^r$. Substituting $a = x^{1/3}$ and $b = \frac{1}{2}x^{-1/3}$ into the expansion of $\left(x^{1/3} + \frac{1}{2x^{1/3}}\right)^{18}$, the general term simplifies to ${}^{18}C_r \left(\frac{1}{2}\right)^r x^{\frac{18-2r}{3}}$. Finding the required coefficients involves setting the exponent of $x$ equal to the specific target values for each term.

For $x^{-2}$, we solve $\frac{18-2r}{3} = -2$, which yields $r = 12$, making the coefficient $m = {}^{18}C_{12} \left(\frac{1}{2}\right)^{12}$. Similarly, for $x^{-4}$, we equate $\frac{18-2r}{3} = -4$, leading to $r = 15$, which gives $n = {}^{18}C_{15} \left(\frac{1}{2}\right)^{15}$.

The ratio is expressed as follows:
$\frac{m}{n} = \frac{{}^{18}C_{12} \left(\frac{1}{2}\right)^{12}}{{}^{18}C_{15} \left(\frac{1}{2}\right)^{15}}$
By simplifying the powers of $\frac{1}{2}$, we obtain $\frac{{}^{18}C_{12}}{{}^{18}C_{15}} \cdot 2^3$. Applying the symmetry property of combinations where ${}^{n}C_r = {}^{n}C_{n-r}$, we find ${}^{18}C_{12} = {}^{18}C_6$ and ${}^{18}C_{15} = {}^{18}C_3$. Calculating the final value results in $\frac{{}^{18}C_6}{{}^{18}C_3} \cdot 8$, which evaluates to 182.

The number of terms in the expansion of a trinomial raised to the power $n$ is determined by the formula $\frac{(n+1)(n+2)}{2}$. Equating this to the given 28 results in the equation $(n+1)(n+2) = 56$. Solving this quadratic relationship reveals that $n=6$, as the product of the consecutive integers 7 and 8 fulfills the requirement.

Finding the sum of all coefficients in an algebraic expansion is achieved by substituting 1 for the variable $x$, as the sum of the coefficients of a polynomial $P(x)$ is simply $P(1)$. Replacing $x$ with 1 in the original expression $\left(1-\frac{2}{x}+\frac{4}{x^2}\right)^n$ yields the base $(1-2+4)^n$. Given that $n=6$, this expression simplifies to $3^6$, which evaluates to 729.

This series acts as a geometric progression where the first term is $(1+x)^{2016}$, the common ratio is $\frac{x}{1+x}$, and the total number of terms is $2017$. Applying the summation formula $S_n = a \frac{1-r^n}{1-r}$ results in $(1+x)^{2016} \cdot \frac{1 - (\frac{x}{1+x})^{2017}}{1 - \frac{x}{1+x}}$. Because the denominator simplifies to $\frac{1}{1+x}$, the entire expression reduces to $(1+x)^{2017} - x^{2017}$. Extracting the coefficient $a_{17}$ of the $x^{17}$ term from the binomial expansion of $(1+x)^{2017}$ yields $\binom{2017}{17}$, which is equivalent to $\frac{2017!}{17!\,2000!}$.