Binomial Theorem – JEE Main MathematicsPractice Questions & PYQs

The given sequence is a finite geometric series with ten terms, where the first term is $a = \frac{1}{2 \cdot 3^{10}}$ and the common ratio is $r = \frac{3}{2}$. Applying the summation formula $S_n = a \frac{r^n - 1}{r - 1}$, we calculate the total sum:
$S = \frac{1}{2 \cdot 3^{10}} \cdot \frac{(3/2)^{10} - 1}{3/2 - 1}$
Since the denominator $3/2 - 1$ equals $1/2$, this term cancels the factor of $2$ in the initial expression's denominator, simplifying the sum to $\frac{(3/2)^{10} - 1}{3^{10}}$. Distributing $3^{10}$ in the denominator yields $\frac{3^{10}/2^{10} - 1}{3^{10}} = \frac{3^{10} - 2^{10}}{2^{10} \cdot 3^{10}}$, which identifies $K = 3^{10} - 2^{10}$.

To find the remainder when $K$ is divided by $6$, we evaluate $3^{10} - 2^{10} \pm od 6$. Every positive integer power of $3$ leaves a remainder of $3$ when divided by $6$, while powers of $2$ alternate between $2$ and $4$ modulo $6$ for odd and even exponents, respectively. Since $10$ is an even exponent, $2^{10} \equiv 4 \pm od 6$, leading to $K \equiv 3 - 4 = -1 \equiv 5 \pm od 6$.

The property $k \cdot {}^nC_k = n \cdot {}^{n-1}C_{k-1}$ allows us to simplify the term $(k \cdot {}^{10}C_k)^2$ as $100 \cdot ({}^9C_{k-1})^2$. Summing this from $k=1$ to $10$ is equivalent to $100 \sum_{k=1}^{10} ({}^9C_{k-1})^2$, which shifts to the following form:

$100 \sum_{j=0}^{9} ({}^9C_j)^2$

Using the identity $\sum_{j=0}^n ({}^nC_j)^2 = {}^{2n}C_n$ with $n=9$, the summation simplifies to $100 \cdot {}^{18}C_9$, which equates to $22000 L$. Solving for $L$ gives $L = 221$.

The binomial expansion of $(1+x)^{10}$ is given by $\sum_{r=0}^{10} C_r x^r$. To determine the sum on the left-hand side, consider the expression for the $r$-th term, which corresponds to $(2r^2-r) C_r$ for $r$ from $1$ to $10$. This can be written as $2 \sum r^2 C_r - \sum r C_r$. Using the standard results $\sum_{r=1}^{10} r C_r = 10 \cdot 2^9$ and $\sum_{r=1}^{10} r^2 C_r = 10 \cdot 11 \cdot 2^8 = 55 \cdot 2^9$, the left-hand side sum becomes $2(55 \cdot 2^9) - 10 \cdot 2^9 = 110 \cdot 2^9 - 10 \cdot 2^9 = 100 \cdot 2^9$, which equals $51200$.

For the right-hand side, the sum $C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \dots$ can be derived by integrating the expansion $(1+x)^{10}$ from $0$ to $1$: $\int_{0}^{1} (1+x)^{10} \, dx = \int_{0}^{1} \sum_{r=0}^{10} C_r x^r \, dx = \sum_{r=0}^{10} \frac{C_r}{r+1}$. This evaluates to $\left[ \frac{(1+x)^{11}}{11} \right]_0^1 = \frac{2^{11}-1}{11}$.

Equating these results within the given structure $100 \cdot 2^9 = \frac{\alpha \cdot 2^{11}}{2^\beta-1} \cdot \left( \frac{2^{11}-1}{11} \right)$, we can simplify the expression. Rearranging the terms yields $51200 = \frac{\alpha \cdot 2048}{2^\beta-1} \cdot \frac{2047}{11}$. Given that $2048 = 2^{11}$ and $2047 = 2^{11}-1$, the equation simplifies to $51200 = \frac{\alpha \cdot 2048}{11}$. Solving for $\alpha$ gives $\alpha = \frac{51200 \times 11}{2048} = 25 \times 11 = 275$. Since $\beta$ corresponds to the power in $2^\beta-1$, we identify $\beta = 11$. Therefore, the value of $\alpha + \beta$ is $275 + 11 = 286$.

The foundation of this problem lies in the fundamental properties of binomial coefficients, specifically the identities where the sum of all binomial coefficients $\sum_{k=0}^{N} {}^{N}C_k$ equals $2^{N}$, and the sum of even-indexed binomial coefficients equals $2^{N-1}$.

For the first factor, we know that ${}^{49}C_0 + {}^{49}C_1 + \dots + {}^{49}C_{49} = 2^{49}$. By isolating the terms from ${}^{49}C_1$ to ${}^{49}C_{49}$, we find that ${}^{49}C_1 + \dots + {}^{49}C_{49} = 2^{49} - 1$. Adding 2 to this value, the first factor simplifies to $2 + 2^{49} - 1$, which is $2^{49} + 1$.

For the second factor, we examine the sum of even-indexed coefficients for $N=50$, which is ${}^{50}C_0 + {}^{50}C_2 + \dots + {}^{50}C_{50} = 2^{49}$. Since ${}^{50}C_0 = 1$, subtracting this from the total yields the required sum, ${}^{50}C_2 + \dots + {}^{50}C_{50} = 2^{49} - 1$.

Substituting these simplified expressions back into the original equation gives $1 + (2^{49} + 1)(2^{49} - 1)$. Applying the difference of squares identity, this expression becomes $1 + (2^{98} - 1)$, which simplifies to $2^{98}$. Since this is equal to $2^{n} \cdot m$ where $m$ is odd, we identify $n=98$ and $m=1$, resulting in a final value of $n+m=99$.

The expansion of $(1+x)^p(1-x)^q$ follows from the product of the individual binomial series, where the coefficient of any term $x^n$ is found by collecting all combinations of terms from the two binomial expansions that multiply to $x^n$. For the linear term, we extract $px$ from the first series and $-qx$ from the second, which gives the relation $p - q = -3$. Extending this to the quadratic term, the coefficient is composed of $\frac{p(p-1)}{2} - pq + \frac{q(q-1)}{2}$, which must equal $-5$. Substituting $p = q - 3$ into this quadratic equation simplifies the expression to $-q + 6 = -5$, identifying the values as $q = 11$ and $p = 8$.

Knowing that the expression is $(1+x)^8(1-x)^{11}$, we determine the coefficient of $x^3$ by summing the four distinct products that yield this power: the constant from $(1+x)^8$ times the $x^3$ term from $(1-x)^{11}$, the $x^1$ term times the $x^2$ term, the $x^2$ term times the $x^1$ term, and the $x^3$ term times the constant. Calculating these components yields $-{}^{11}C_{3} + {}^{8}C_{1} \cdot {}^{11}C_{2} - {}^{8}C_{2} \cdot {}^{11}C_{1} + {}^{8}C_{3}$, which corresponds to $-165 + 8(55) - 28(11) + 56$. Summing these values gives $-165 + 440 - 308 + 56$, which equals 23.

Finding the constant term in a trinomial expansion is best handled by applying the multinomial theorem, which identifies coefficients based on specific power distributions. For the expression $(3 x^3 - 2 x^2 + 5x^{-5})^{10}$, finding the constant term is equivalent to isolating the coefficient of $x^{50}$ within the expansion of $(3 x^8 - 2 x^7 + 5)^{10}$ after factoring $x^{-50}$ from the original base. The general term for this expansion takes the form $\frac{10!}{n_1! n_2! n_3!} (3 x^8)^{n_1} (-2 x^7)^{n_2} (5)^{n_3}$, where the sum of indices $n_1 + n_2 + n_3$ must equal 10.

We require the exponent of $x$ to be 50, which leads to the constraint $8n_1 + 7n_2 = 50$. By inspection, the only valid set of non-negative integers satisfying this condition alongside the sum constraint is $n_1 = 1$, $n_2 = 6$, and $n_3 = 3$. Inserting these values into the multinomial coefficient expression, we calculate the coefficient as $\frac{10!}{1! 6! 3!} (3)^1 (-2)^6 (5)^3$. Extracting the powers of 2 from this product involves recognizing that $\frac{10 \times 9 \times 8 \times 7}{6} = 840 = 2^3 \times 105$, which when multiplied by $(-2)^6 = 2^6$ results in $2^9$ multiplied by an odd integer. Thus, the value of $k$ is 9.

The binomial theorem effectively expands the given polynomials, allowing us to isolate $\alpha$ and $\beta$. By writing $9^n$ as $(1+8)^n$, the binomial expansion $1 + \binom{n}{1}8 + \binom{n}{2}8^2 + \binom{n}{3}8^3 + \dots + \binom{n}{n}8^n$ allows us to subtract the linear terms $1+8n$ and divide the result by 64 to obtain $\alpha = \binom{n}{2} + \binom{n}{3}8 + \binom{n}{4}8^2 + \dots + \binom{n}{n}8^{n-2}$. Applying an identical procedure to $6^n-5n-1$ and dividing by 25 yields $\beta = \binom{n}{2} + \binom{n}{3}5 + \binom{n}{4}5^2 + \dots + \binom{n}{n}5^{n-2}$. Subtracting these two expressions causes the common $\binom{n}{2}$ terms to vanish, leaving $\binom{n}{3}(8-5) + \binom{n}{4}(8^2-5^2) + \dots + \binom{n}{n}(8^{n-2}-5^{n-2})$.

Modular arithmetic simplifies finding the remainder of exponential expressions by focusing on the base's behavior relative to the divisor. Since we are dividing by 5, we can observe that $7^2 = 49$ and $3^2 = 9$ are both one less than a multiple of 5, meaning they are each equivalent to $-1$ modulo 5. The expression $7^{2022} + 3^{2022}$ can therefore be rewritten as $(7^2)^{1011} + (3^2)^{1011}$. Replacing these bases yields $(-1)^{1011} + (-1)^{1011}$, which simplifies to $-1 + (-1) = -2$. Adding the divisor 5 to this negative result gives the positive remainder $5 - 2 = 3$.

The first summation $\sum_{k=1}^{31} {}^{31}C_k \cdot {}^{31}C_{k-1}$ can be re-indexed using the symmetry property ${}^{n}C_r = {}^{n}C_{n-r}$, which transforms it into $\sum_{k=1}^{31} {}^{31}C_k \cdot {}^{31}C_{32-k}$. This sum represents the coefficient of $x^{32}$ in the binomial expansion of $(1+x)^{31}(1+x)^{31}$, or $(1+x)^{62}$, which simplifies to ${}^{62}C_{32}$ or ${}^{62}C_{30}$. Similarly, the second summation $\sum_{k=1}^{30} {}^{30}C_k \cdot {}^{30}C_{k-1}$ corresponds to the coefficient of $x^{31}$ in $(1+x)^{30}(1+x)^{30}$, or $(1+x)^{60}$, which simplifies to ${}^{60}C_{31}$ or ${}^{60}C_{29}$.

The problem provides the equation ${}^{62}C_{30} - {}^{60}C_{29} = \frac{\alpha(60!)}{30!31!}$. Expanding these combinations into factorial form gives $\frac{62!}{30!32!} - \frac{60!}{29!31!} = \frac{\alpha(60!)}{30!31!}$. By multiplying the entire equation by $\frac{30!31!}{60!}$, the terms become $\frac{62 \cdot 61 \cdot 60!}{30! \cdot 32 \cdot 31!} \cdot \frac{30!31!}{60!} - \frac{60!}{29! \cdot 31!} \cdot \frac{30!31!}{60!} = \alpha$.

Simplifying these fractions results in $\frac{62 \cdot 61}{32} - 30 = \alpha$. To find the value of $16\alpha$, distribute 16 into the expression to yield $16 \left( \frac{62 \cdot 61}{32} \right) - 16(30)$, which simplifies to $\frac{62 \cdot 61}{2} - 480$. This further reduces to $31 \cdot 61 - 480$, resulting in $1891 - 480$, which equals 1411.

Identifying the constant term in a binomial expansion involves finding the specific term where the variable carries an exponent of zero. For the expression $(a x^{1/8} + b x^{-1/12})^{10}$, the general term $T_{r+1}$ is defined by ${}^{10}C_r (a x^{1/8})^{10-r} (b x^{-1/12})^r$. Setting the aggregate exponent of $x$, calculated as $\frac{10-r}{8} - \frac{r}{12}$, equal to zero leads to $r=6$. This result transforms the general term into ${}^{10}C_6 a^4 b^6$, which simplifies to $210 a^4 b^6$.

Minimizing this constant term given the constraint $\frac{1}{a^2} + \frac{1}{b^3} = 4$ is achieved by utilizing the inequality between the geometric mean and the harmonic mean. The relationship $\sqrt{u \cdot v} \geq \frac{2}{\frac{1}{u} + \frac{1}{v}}$ applied with $u=a^2$ and $v=b^3$ yields $\sqrt{a^2 b^3} \geq \frac{2}{4}$, simplifying to $\sqrt{a^2 b^3} \geq \frac{1}{2}$. Squaring both sides of this inequality produces $a^2 b^3 \geq \frac{1}{4}$, which implies $a^4 b^6 \geq \frac{1}{16}$. Multiplying the coefficient $210$ by this minimum value of $\frac{1}{16}$ results in $\frac{105}{8}$.

The binomial theorem allows us to represent the expansion of $\left(2 x^3 + \frac{3}{x}\right)^{10}$ through a general term $T_{r+1} = {}^{10}C_r (2x^3)^{10-r} \left(\frac{3}{x}\right)^r$, which simplifies to ${}^{10}C_r \cdot 2^{10-r} \cdot 3^r \cdot x^{30-4r}$. For the coefficient to correspond to a positive even power of $x$, the exponent $30-4r$ must be both greater than zero and an even number. Since $30$ and $4r$ are both even, the expression $30-4r$ is inherently even, so we only need to satisfy the condition $30-4r > 0$, or $r < 7.5$. This implies the valid range for $r$ is the set of integers from $0$ to $7$.

Finding the sum of coefficients for $r=0$ to $7$ is most effectively done by taking the total sum of all coefficients in the expansion and subtracting the values for the remaining terms where $r=8, 9,$ and $10$. Setting $x=1$ in the original expression yields the total sum $(2+3)^{10} = 5^{10}$. The excluded terms are calculated as follows: for $r=8$, the term is ${}^{10}C_8 \cdot 2^2 \cdot 3^8 = 45 \cdot 4 \cdot 3^8 = 60 \cdot 3^9$; for $r=9$, the term is ${}^{10}C_9 \cdot 2^1 \cdot 3^9 = 10 \cdot 2 \cdot 3^9 = 20 \cdot 3^9$; and for $r=10$, the term is ${}^{10}C_{10} \cdot 2^0 \cdot 3^{10} = 3 \cdot 3^9$. Combining these gives a total subtraction of $(60+20+3) \cdot 3^9 = 83 \cdot 3^9$. This confirms the total sum is $5^{10} - 83 \cdot 3^9$, leading to $\beta = 83$.

In the expansion of $\left(3^{\frac{1}{4}}+5^{\frac{1}{8}}\right)^{60}$, a term $T_{r+1}={}^{60}C_{r}3^{\frac{r}{4}}5^{\frac{60-r}{8}}$ is rational only when the exponents $\frac{r}{4}$ and $\frac{60-r}{8}$ are integers. This requires $r$ to be a multiple of 4, while simultaneously ensuring that $60-r$ is a multiple of 8. Testing the multiples of 4 within the range $0 \le r \le 60$ against the condition that $60-r$ must also be divisible by 8 yields the set $\{4, 12, 20, 28, 36, 44, 52, 60\}$. There are exactly 8 such values for $r$, which correspond to the 8 rational terms in the expansion. Since there are $60+1=61$ total terms, the number of irrational terms is $n = 61 - 8 = 53$. Consequently, $n-1 = 52$, a value divisible by 26.

Modular arithmetic allows us to simplify large base operations by focusing on remainders. When 2021 is divided by 7, it leaves a remainder of 5, which can be treated as $-2$ to make calculations easier, and 2022 divided by 7 leaves a remainder of 6, which is equivalent to $-1$. Substituting these into the expression changes the problem to $(-2)^{2022} + (-1)^{2021}$.

Since the exponent 2022 is even, the negative base becomes positive, turning $(-2)^{2022}$ into $2^{2022}$. By noting that $2^3 = 8$ is congruent to $1 \pm od{7}$, and seeing that 2022 is a multiple of 3, we can express this as $(2^3)^{674}$, which simplifies to $1^{674}$, resulting in $1$. Simultaneously, the odd exponent on $-1$ preserves the negative sign, so $(-1)^{2021}$ equals $-1$. Adding these two terms together yields $1 + (-1) = 0$, demonstrating that the original expression is a multiple of 7 and leaves no remainder.

The expansion $(a+2b+4ab)^{10}$ follows the multinomial theorem, where the general term is defined as $\frac{10!}{n_1! n_2! n_3!} a^{n_1} (2b)^{n_2} (4ab)^{n_3}$. We are interested in the term containing $a^7 b^8$, which implies $a^{n_1+n_3} b^{n_2+n_3} = a^7 b^8$. Equating the exponents and knowing the sum of terms $n_1+n_2+n_3$ must equal $10$, we solve the system $n_1+n_3=7$, $n_2+n_3=8$, and $n_1+n_2+n_3=10$. Subtracting the first two equations from the sum shows $n_3=5$, which further reveals $n_1=2$ and $n_2=3$. Plugging these values into the general term gives $\frac{10!}{2! 3! 5!} \cdot 1^2 \cdot 2^3 \cdot 4^5$. Expressing $4^5$ as $2^{10}$ combines the powers of 2 into $2^{13}$, resulting in the coefficient $\frac{10!}{2! 3! 5!} \cdot 2^{13}$. Calculating the factorial expression yields $2520 \cdot 2^{13}$. Since $2520$ is equivalent to $315 \cdot 8$, or $315 \cdot 2^3$, we express the total coefficient as $315 \cdot 2^3 \cdot 2^{13}$, which equals $315 \cdot 2^{16}$. Therefore, $k$ is $315$.

Simplifying the bracketed expression is the most direct approach to identifying the independent term. The first fraction contains a sum of cubes, which simplifies the entire term to $x^{1/3} + 1$. Factoring the numerator and denominator of the second fraction as $(\sqrt{x}-1)(\sqrt{x}+1)$ and $\sqrt{x}(\sqrt{x}-1)$ respectively allows it to be reduced to $1 + x^{-1/2}$. Subtracting these two components eliminates the constant values, condensing the original expression to $(x^{1/3} - x^{-1/2})^{10}$

The binomial theorem provides the general term $T_{r+1} = {}^{10}C_r (x^{1/3})^{10-r} (-x^{-1/2})^r$, which allows us to isolate the variable powers. Combining these gives a total exponent of $\frac{10-r}{3} - \frac{r}{2}$ Since the term must be independent of $x$, this exponent must equal zero, leading to the equation $20 - 5r = 0$, which yields $r=4$. Calculating the coefficient for this term results in 210.

The binomial theorem dictates that any term in the expansion of a binomial expression $(A+B)^n$ is defined by the general term formula
$T_{r+1} = {}^nC_r A^{n-r} B^r$
For the expression $\left(x \sin \alpha + \frac{a \cos \alpha}{x}\right)^{10}$, the general term is ${}^{10}C_r (x \sin \alpha)^{10-r} \left( \frac{a \cos \alpha}{x} \right)^r$. Extracting the powers of $x$ from this expansion yields $x^{10-r} \cdot x^{-r} = x^{10-2r}$. To obtain a term independent of $x$, the variable must be eliminated, meaning the exponent must equal zero; setting $10-2r = 0$ identifies $r=5$ as the position of the constant term.

Substituting $r=5$ into the expansion results in the constant value
$T_{6} = {}^{10}C_5 (\sin \alpha)^5 (a \cos \alpha)^5 = {}^{10}C_5 a^5 (\sin \alpha \cos \alpha)^5$
Utilizing the trigonometric identity $\sin \alpha \cos \alpha = \frac{1}{2} \sin 2\alpha$, the expression simplifies further to ${}^{10}C_5 a^5 \left( \frac{\sin 2\alpha}{2} \right)^5$, or ${}^{10}C_5 \frac{a^5}{32} (\sin 2\alpha)^5$. Since the sine function reaches a maximum of 1, the greatest possible value for this term occurs when $\sin 2\alpha = 1$. By equating this maximal value to the provided $\frac{10!}{(5!)^2}$, which is equivalent to ${}^{10}C_5$, we obtain the equation
${}^{10}C_5 \frac{a^5}{32} = {}^{10}C_5$
Dividing both sides by ${}^{10}C_5$ leaves $\frac{a^5}{32} = 1$, which confirms that $a^5 = 32$ and thus $a=2$.

For any binomial expansion $(1+x)^n$, the number and position of middle terms depend on whether the exponent $n$ is even or odd. When $n$ is even, a single middle term appears at the position $\frac{n}{2}+1$, whereas an odd $n$ results in two middle terms at positions $\frac{n+1}{2}$ and $\frac{n+3}{2}$. In the case of $(1+x)^{20}$, the single middle term coefficient is identified as ${}^{20}C_{10}$. Conversely, $(1+x)^{19}$ possesses two middle terms with coefficients of ${}^{19}C_{9}$ and ${}^{19}C_{10}$. The sum of these coefficients, ${}^{19}C_{9}+{}^{19}C_{10}$, simplifies via the standard additive identity for binomial coefficients, ${}^{n}C_{r}+{}^{n}C_{r-1}={}^{n+1}C_{r}$, which yields ${}^{20}C_{10}$. Because both the numerator and the denominator are equal to ${}^{20}C_{10}$, the required ratio is 1.

The expansion of $(1-x+x^{3})^{n}$ into a series $\sum_{j=0}^{3n} a_{j} x^{j}$ allows us to determine the properties of its coefficients by substituting specific values for $x$. Setting $x=1$ results in $(1-1+1)^{n} = 1$, which equates to the sum of all coefficients, $a_{0}+a_{1}+a_{2}+\dots+a_{3n} = 1$. By setting $x=-1$, we obtain $(1+1-1)^{n} = 1$, which equals the alternating sum $a_{0}-a_{1}+a_{2}-a_{3}+\dots = 1$.

Adding these two results gives $2(a_{0}+a_{2}+a_{4}+\dots) = 2$, identifying the sum of the even-indexed coefficients as $a_{0}+a_{2}+a_{4}+\dots = 1$. Subtracting the second equation from the first yields $2(a_{1}+a_{3}+a_{5}+\dots) = 0$, identifying the sum of the odd-indexed coefficients as $0$. Applying these results to the expression, we substitute the sum of even-indexed terms and the sum of odd-indexed terms, leading to a final value of $1$.

Modular arithmetic allows for the simplification of large bases relative to the divisor before performing exponentiation. Dividing 2021 by 17 gives a remainder of 15, which is equivalent to $-2 \pm od{17}$. Therefore, the expression $(2021)^{3762}$ is equivalent to $(-2)^{3762} \pm od{17}$, and since the exponent is even, this simplifies to $2^{3762} \pm od{17}$.

We know that $2^4 = 16$, which is congruent to $-1 \pm od{17}$. We can express $2^{3762}$ as $2^2 \times (2^4)^{940}$, which is $4 \times (16)^{940}$. Replacing 16 with $-1$ leads to $4 \times (-1)^{940} \pm od{17}$. Because 940 is an even number, the term $(-1)^{940}$ becomes 1, leaving us with $4 \times 1 \equiv 4 \pm od{17}$. This confirms that the final remainder is 4.

The binomial theorem provides a straightforward way to identify the coefficient of any term in an expansion, expressed by the general form ${}^{n}C_{r} a^{n-r} b^{r}$ for the expression $(a+b)^n$. In the case of $(2+\frac{x}{3})^n$, the coefficient associated with the term containing $x^r$ is ${}^{n}C_{r} \cdot 2^{n-r} \cdot (\frac{1}{3})^r$. By equating the coefficients for $x^7$ and $x^8$, we establish the relation ${}^{n}C_{7} 2^{n-7} \frac{1}{3^7} = {}^{n}C_{8} 2^{n-8} \frac{1}{3^8}$.

To isolate $n$, reorganize the equation by grouping the binomial coefficients on one side and the numerical constants on the other, resulting in $\frac{{}^{n}C_{8}}{{}^{n}C_{7}} = \frac{2^{n-7} \cdot 3^8}{2^{n-8} \cdot 3^7}$. Simplifying the exponents on the right side leaves $2^1 \cdot 3^1$, which equals $6$. Expanding the ratio of the combinations $\frac{{}^{n}C_{8}}{{}^{n}C_{7}}$ produces $\frac{n-8+1}{8}$, simplifying to $\frac{n-7}{8}$. Equating this to $6$ yields $n-7 = 48$, which confirms that the value of $n$ is $55$.