Binomial Theorem – JEE Main MathematicsPractice Questions & PYQs

To find the constant term in the expansion of the product of a trinomial and a binomial power, we must distribute each term of the trinomial $(1+2x-3x^3)$ across the binomial expansion $(\frac{3}{2}x^2 - \frac{1}{3x})^9$ and identify which combinations result in a constant term, which is the coefficient of $x^0$. The general term of the binomial part is given by:

$\binom{9}{r} \left(\frac{3}{2}x^2\right)^{9-r} \left(-\frac{1}{3}x^{-1}\right)^r = \binom{9}{r} \left(\frac{3}{2}\right)^{9-r} \left(-\frac{1}{3}\right)^r x^{18-2r-r} = \binom{9}{r} \left(\frac{3}{2}\right)^{9-r} \left(-\frac{1}{3}\right)^r x^{18-3r}$

To obtain a constant in the final product, we need to multiply the factors of the trinomial by terms in the expansion that produce $x^0$, $x^{-1}$, and $x^{-3}$ respectively. For the $1$ in the trinomial, we look for $18-3r=0$, which gives $r=6$. This provides a term of $\binom{9}{6} (\frac{3}{2})^3 (-\frac{1}{3})^6 = 84 \cdot \frac{27}{8} \cdot \frac{1}{729} = \frac{7}{18}$. For the $2x$ in the trinomial, we require $18-3r=-1$, which has no integer solution for $r$. For the $-3x^3$ in the trinomial, we look for $18-3r=-3$, which gives $r=7$. This provides a term of $\binom{9}{7} (\frac{3}{2})^2 (-\frac{1}{3})^7 = 36 \cdot \frac{9}{4} \cdot (-\frac{1}{2187}) = -\frac{1}{27}$.

Combining these, the constant term $p$ is the sum of the products:

$p = 1 \cdot \left(\frac{7}{18}\right) + 0 + (-3) \cdot \left(-\frac{1}{27}\right) = \frac{7}{18} + \frac{1}{9} = \frac{7+2}{18} = \frac{9}{18} = \frac{1}{2}$

Finally, calculating the requested value gives $108 p = 108 \cdot \frac{1}{2} = 54$.

The relationship between a sequence's terms and its partial sums relies on the identity $a_n = S_n - S_{n-1}$, which allows for the isolation of each term from a given series expression. By rewriting the partial sum $S_n = \frac{n^2 + 3n}{(n+1)(n+2)}$ as $1 - \frac{2}{(n+1)(n+2)}$, finding individual terms becomes straightforward. Subtracting $S_{n-1}$ from $S_n$ yields the general term $a_n = \frac{4}{n(n+1)(n+2)}$, which implies that the reciprocal term in the required summation is $\frac{1}{a_k} = \frac{k(k+1)(k+2)}{4}$.

With the reciprocal defined, the expression $28 \sum_{k=1}^{10} \frac{1}{a_k}$ simplifies to $7 \sum_{k=1}^{10} k(k+1)(k+2)$. This summation follows the standard telescoping identity for the product of consecutive integers, $\sum_{k=1}^n k(k+1)(k+2) = \frac{n(n+1)(n+2)(n+3)}{4}$. Applying this formula for $n=10$, the calculation $7 \times \frac{10 \times 11 \times 12 \times 13}{4}$ reduces to $7 \times 10 \times 11 \times 3 \times 13$. Factoring this result into primes yields $2 \times 3 \times 5 \times 7 \times 11 \times 13$, confirming that the product is composed of the first six prime numbers. Thus, $m=6$.

For the binomial expansion of $(1+2x)^n$, the general term is expressed as $t_{r+1} = {}^n C_r (2x)^r$, which implies the coefficient of the $(r+1)$-th term is ${}^n C_r 2^r$. When considering three consecutive terms with coefficients in the ratio $2:5:8$, we represent these as $C_{r-1} = {}^n C_{r-1} 2^{r-1}$, $C_r = {}^n C_r 2^r$, and $C_{r+1} = {}^n C_{r+1} 2^{r+1}$. By establishing the ratios $\frac{C_{r-1}}{C_r} = \frac{2}{5}$ and $\frac{C_r}{C_{r+1}} = \frac{5}{8}$, we create a system of equations involving $n$ and $r$.

Expanding the first ratio leads to the relationship $\frac{r}{n-r+1} \cdot \frac{1}{2} = \frac{2}{5}$, which simplifies to $9r = 4n + 4$. Similarly, analyzing the second ratio gives $\frac{r+1}{n-r} \cdot \frac{1}{2} = \frac{5}{8}$, simplifying to $9r = 5n - 4$. Since both expressions equal $9r$, setting them equal provides $4n + 4 = 5n - 4$, revealing that $n=8$. Substituting this value back into either equation yields $9r = 36$, or $r=4$.

With $n=8$ and $r=4$, the middle term's coefficient is calculated as ${}^8 C_4 2^4$. Evaluating the binomial coefficient ${}^8 C_4$ gives $70$, which, when multiplied by $16$ (since $2^4 = 16$), results in a value of 1120.

Utilizing the symmetry property of binomial coefficients, where ${}^{n}C_k = {}^{n}C_{n-k}$, the sum $S = \sum_{k=0}^{30} k({}^{30}C_k)^2$ can be paired with its equivalent form $\sum_{k=0}^{30} (30-k)({}^{30}C_{30-k})^2$ to show that $2S = 30 \sum_{k=0}^{30} ({}^{30}C_k)^2$. Applying the standard identity $\sum_{k=0}^{n} ({}^{n}C_k)^2 = {}^{2n}C_n$ with $n=30$ reduces this to $2S = 30 \cdot {}^{60}C_{30}$, which simplifies to $S = 15 \cdot {}^{60}C_{30}$. Substituting the factorial definition ${}^{60}C_{30} = \frac{60!}{(30!)^2}$ results in $S = 15 \cdot \frac{60!}{(30!)^2}$, directly establishing that $\alpha = 15$.

The binomial theorem provides the general term for the expansion $\left( \frac{x^{5/2}}{2} - \frac{4}{x^{\ell}} \right)^9$ as $T_{r+1} = \binom{9}{r} \left( \frac{x^{5/2}}{2} \right)^{9-r} \left( -\frac{4}{x^{\ell}} \right)^r$. By simplifying the expression, we obtain $T_{r+1} = \binom{9}{r} (-1)^r 2^{3r-9} x^{\frac{45 - r(5+2\ell)}{2}}$. The constant term occurs when the exponent of $x$ is zero, implying $45 - r(5+2\ell) = 0$, while the numerical value is given as $-84$. Since $\binom{9}{3} = 84$, setting $r=3$ yields $(-1)^3 \binom{9}{3} 2^{3(3)-9} = -84$, which confirms $r=3$. Substituting this value into the exponent equation $45 - 3(5+2\ell) = 0$ reveals that $5+2\ell = 15$, leading to $\ell=5$.

We then determine the coefficient of $x^{-3\ell}$, which corresponds to $x^{-15}$. By setting the exponent $\frac{45 - r(5+2(5))}{2} = -15$, we simplify to $45 - 15r = -30$, resulting in $r=5$. The coefficient for this term is $\binom{9}{5} (-1)^5 2^{3(5)-9} = -126 \cdot 2^6 = -63 \cdot 2^7$. Comparing this with the expression $2^{\alpha} \beta$, we identify $\alpha = 7$ and $\beta = -63$, where $\beta$ is indeed an odd negative number. Evaluating the final expression $|\alpha \ell - \beta|$ yields $|7(5) - (-63)|$, which equals $98$.

Using the multinomial expansion for $(1 - x + 2x^3)^{10}$, the general term follows the form $\frac{10!}{a!b!c!} (1)^a (-x)^b (2x^3)^c$, which simplifies to $\frac{10!}{a!b!c!} (-1)^b (2)^c x^{b+3c}$. We identify the terms containing $x^7$ by solving the system $a + b + c = 10$ and $b + 3c = 7$ for non-negative integers.

Setting $c = 0$ yields $b = 7$ and $a = 3$, producing a coefficient of $\frac{10!}{3!7!0!} (-1)^7 = -120$. Setting $c = 1$ leads to $b = 4$ and $a = 5$, giving $\frac{10!}{5!4!1!} (-1)^4 (2)^1 = 2520$. Setting $c = 2$ results in $b = 1$ and $a = 7$, contributing $\frac{10!}{7!1!2!} (-1)^1 (2)^2 = -1440$. Combining these values, the final coefficient is $-120 + 2520 - 1440$, which equals 960.

To determine the divisibility of $25^{190}-19^{190}-8^{190}+2^{190}$, we can regroup the expression as $(25^{190}-8^{190})-(19^{190}-2^{190})$. Based on the principle that the difference of two powers $a^n - b^n$ is always divisible by $a-b$, we can identify that $25^{190}-8^{190}$ is divisible by $25-8=17$, and $19^{190}-2^{190}$ is divisible by $19-2=17$. Since both components are divisible by 17, the entire expression must be divisible by 17.

Because $25^{190}$ and $19^{190}$ are odd, while $8^{190}$ and $2^{190}$ are even, the overall result of the expression is an even number, which provides an additional factor of 2. Combining these, the expression is divisible by $17 \times 2 = 34$. To check for divisibility by 14, we would require the expression to be divisible by both 2 and 7; however, evaluating the terms modulo 7 demonstrates that the expression leaves a non-zero remainder, confirming that it is not divisible by 7 and, consequently, not divisible by 14.

To find the term independent of x in a binomial expansion, the general term formula $T_{r+1} = {}^{n}C_r a^{n-r} b^r$ is utilized, where the variable part is isolated to identify when its net exponent becomes zero. For the expression $\left(x^{\frac{2}{3}}+\frac{\alpha}{x^{3}}\right)^{22}$, the general term is represented as:
$T_{r+1} = {}^{22}C_r \cdot (x^{\frac{2}{3}})^{22-r} \cdot (\alpha x^{-3})^r$
Consolidating the powers of x, the expression becomes $x^{\frac{44-2r}{3}} \cdot x^{-3r}$, which simplifies to $x^{\frac{44-2r-9r}{3}} = x^{\frac{44-11r}{3}}$. For the term to be independent of x, the exponent must equal 0, leading to the condition $44 - 11r = 0$, which yields $r = 4$. Substituting this value back into the coefficient portion of the general term results in:
${}^{22}C_4 \cdot \alpha^4 = 7315$
Evaluating the binomial coefficient ${}^{22}C_4$ as $\frac{22 \times 21 \times 20 \times 19}{4 \times 3 \times 2 \times 1}$ yields 7315. The equation then simplifies to $7315 \cdot \alpha^4 = 7315$, which implies $\alpha^4 = 1$. Taking the fourth root, the absolute value $|\alpha|$ is 1.

To determine the maximum value of a product where variables have specific exponents and a fixed sum, the most effective approach is to apply the Arithmetic Mean-Geometric Mean inequality. We distribute the sum $a+b+c+d=11$ into individual terms that correspond to the exponents of the expression $a^5 b^3 c^2 d$, effectively splitting $a$ into five parts of $\frac{a}{5}$, $b$ into three parts of $\frac{b}{3}$, $c$ into two parts of $\frac{c}{2}$, and $d$ into one part of $d$. Summing these $5+3+2+1=11$ components results in $a+b+c+d=11$.

Applying the inequality to these $11$ terms states that their arithmetic mean is greater than or equal to their geometric mean, which can be expressed as $\frac{5(\frac{a}{5}) + 3(\frac{b}{3}) + 2(\frac{c}{2}) + d}{11} \geq \sqrt[11]{(\frac{a}{5})^5 (\frac{b}{3})^3 (\frac{c}{2})^2 d}$. Given the sum is $11$, the left side becomes $1$, leading to the inequality $1 \geq \sqrt[11]{\frac{a^5 b^3 c^2 d}{5^5 \cdot 3^3 \cdot 2^2}}$. By raising both sides to the power of $11$, we find that $a^5 b^3 c^2 d \leq 5^5 \cdot 3^3 \cdot 2^2$. Calculating this product yields $3125 \cdot 27 \cdot 4 = 337500$. Since the maximum value is given as $3750 \beta$, we set $3750 \beta = 337500$, which simplifies to $\beta = 90$.

A sequence in geometric progression follows the rule $a_n = ar^{n-1}$. For the third and fifth terms, their product is $ar^2 \cdot ar^4 = a^2r^6$, which is the square of the fourth term $a_4^2$. Since this product is $\frac{1}{9}$, the fourth term $a_4 = ar^3$ must be $\frac{1}{3}$.

Applying the condition $a_6 + a_8 = 2$, we rewrite the sum as $ar^5 + ar^7 = ar^3(r^2 + r^4) = 2$. Substituting the value $ar^3 = \frac{1}{3}$ yields $\frac{1}{3}(r^2 + r^4) = 2$, which simplifies to $r^4 + r^2 - 6 = 0$. Factoring this quadratic in $r^2$ reveals $(r^2+3)(r^2-2) = 0$, and because the sequence consists of increasing positive terms, $r^2 = 2$ is the only valid solution.

To determine the value of $6(a_2+a_4)(a_4+a_6)$, we express these sums in terms of $a_4$ and $r$ as $a_4(\frac{1}{r^2} + 1)$ and $a_4(1 + r^2)$. Substituting $a_4 = \frac{1}{3}$ and $r^2 = 2$ leads to:
$6 \cdot \left(\frac{1}{3}\left(\frac{1}{2} + 1\right)\right) \cdot \left(\frac{1}{3}(1 + 2)\right) = 6 \cdot \left(\frac{1}{2}\right) \cdot (1) = 3$

The series provided is a higher-order arithmetico-geometric series, where the difference between successive terms requires repeated shifting and subtraction to reduce the expression to a standard geometric progression. Setting the sum to $S=10$, we subtract the first term to isolate the remaining infinite series: $9 = \frac{4}{k} + \frac{8}{k^2} + \frac{13}{k^3} + \frac{19}{k^4} + \ldots$. Multiplying this equation by $\frac{1}{k}$ and subtracting the result from the previous line simplifies the series coefficients, yielding $9(1 - \frac{1}{k}) = \frac{4}{k} + \frac{4}{k^2} + \frac{5}{k^3} + \frac{6}{k^4} + \ldots$.

Applying this reduction technique a second time by multiplying the new expression by $\frac{1}{k}$ and subtracting it again leaves us with $9(1 - \frac{1}{k})^2 = \frac{4}{k} + (\frac{1}{k^3} + \frac{1}{k^4} + \frac{1}{k^5} + \ldots)$. The terms inside the parenthesis form an infinite geometric series with the first term $\frac{1}{k^3}$ and common ratio $\frac{1}{k}$, which sums to $\frac{1/k^3}{1 - 1/k} = \frac{1}{k^2(k-1)}$. This gives us the equation $9(1 - \frac{1}{k})^2 = \frac{4}{k} + \frac{1}{k^2(k-1)}$. Multiplying the entire equation by $k^2(k-1)$ leads to $9(k-1)^3 = 4k(k-1) + 1$. Solving this cubic equation for natural numbers reveals that $k=2$ is the unique solution.

Determining a specific coefficient in a binomial expansion relies on the general term formula $T_{r+1} = {}^nC_r a^{n-r} b^r$, which allows us to isolate the variable component effectively. For the expansion $(\frac{4x}{5} + \frac{5}{2x^2})^9$, the general term simplifies to the expression ${}^9C_r (\frac{4}{5})^{9-r} (\frac{5}{2})^r x^{9-3r}$. Setting the exponent $9-3r = -6$ yields $3r = 15$, or $r = 5$. By substituting $r = 5$ into the general term, we compute the coefficient as ${}^9C_5 (\frac{4}{5})^4 (\frac{5}{2})^5$, which evaluates to $126 \cdot \frac{256}{625} \cdot \frac{3125}{32}$ and results in 5040.

A geometric progression defined by terms $a$, $ar$, and $ar^2$ has a sum of squares that simplifies to the following expression:

$a^2(1 + r^2 + r^4) = 33033$

Factoring the value 33033 into $121 \times 273$ reveals that the first term squared is $a^2 = 121$, which means $a = 11$. Consequently, the remaining factor satisfies the equation $1 + r^2 + r^4 = 273$.

Isolating the common ratio requires rewriting the equation as $r^2(1 + r^2) = 272$. Since 272 factors into $16 \times 17$, it follows that $r^2 = 16$, identifying the common ratio as $r = 4$. Calculating the sum of the three terms $a(1 + r + r^2)$ by substituting these values gives $11(1 + 4 + 16)$, which simplifies to $11 \times 21$, resulting in 231.

Determining the remainder of $(2023)^{2023}$ divided by $35$ becomes straightforward when we decompose the divisor into its prime factors, 5 and 7, allowing us to evaluate the expression modulo each factor. Because 2023 is a multiple of 7, it follows that $(2023)^{2023} \equiv 0 \pm od{7}$. Turning to the modulus 5, note that $2023 \equiv 3 \pm od{5}$, which simplifies the task to finding $3^{2023} \pm od{5}$.

The powers of 3 modulo 5 follow a cycle of length 4: $3^1 \equiv 3$, $3^2 \equiv 4$, $3^3 \equiv 2$, and $3^4 \equiv 1$. Dividing the exponent 2023 by this cycle length 4 leaves a remainder of 3, meaning $3^{2023} \equiv 3^3 \equiv 2 \pm od{5}$. We now search for a value $x$ satisfying the system $x \equiv 0 \pm od{7}$ and $x \equiv 2 \pm od{5}$. Since 7 is the smallest multiple of 7 that leaves a remainder of 2 when divided by 5, the final remainder is 7.

Finding the remainder of a large sum modulo a number is best handled by reducing the base terms through modular congruence. Given that the 50th root of $x$ is 12 and the 50th root of $y$ is 18, the values are $x = 12^{50}$ and $y = 18^{50}$. These expressions can be rewritten as $x = (12^2)^{25} = 144^{25}$ and $y = (18^2)^{25} = 324^{25}$. Working modulo 25, we observe that $144 \equiv -6 \pm od{25}$ and $324 \equiv -1 \pm od{25}$, so the sum $x+y$ becomes congruent to $(-6)^{25} + (-1)^{25} \pm od{25}$, which simplifies to $-(6^{25} + 1)$.

The evaluation of $6^{25} \pm od{25}$ relies on treating $6$ as $(1+5)$ and applying the binomial theorem: $(1+5)^{25} = 1 + \binom{25}{1}5 + \binom{25}{2}5^2 + \dots$. Since $\binom{25}{1}5 = 125$ is a multiple of 25 and every subsequent term contains at least $5^2$, all terms except the first one vanish under modulo 25, leaving $6^{25} \equiv 1 \pm od{25}$. Substituting this back, the expression $-(6^{25} + 1)$ becomes $-(1 + 1) = -2 \pm od{25}$. Adding 25 to the negative result of $-2$ yields a final remainder of 23.

Multiplying the entire infinite series $P$ by the factor $(\frac{1}{2} + \frac{1}{3})$ is an effective strategy to simplify the expression. This algebraic maneuver leverages the relationship between consecutive terms, causing the cross-products to cancel out and collapsing the infinite sum into two cleaner, manageable geometric series.

Performing this multiplication results in the equality $\frac{5P}{6} = (\frac{1}{2^2} - \frac{1}{3^2}) + (\frac{1}{2^3} + \frac{1}{3^3}) + (\frac{1}{2^4} - \frac{1}{3^4}) + \dots$. By separating the terms based on their denominators, we can evaluate the infinite sums using the standard formula $S = \frac{a}{1-r}$. This approach yields $\frac{\frac{1}{4}}{1 - \frac{1}{2}} - \frac{\frac{1}{9}}{1 + \frac{1}{3}}$, which simplifies to $\frac{1}{2} - \frac{1}{12} = \frac{5}{12}$.

Equating $\frac{5P}{6}$ to $\frac{5}{12}$ identifies the value of $P$ as $\frac{1}{2}$. Given that $P = \frac{\alpha}{\beta}$ with $\alpha = 1$ and $\beta = 2$, calculating the requested expression $\alpha + 3\beta$ results in $1 + 3(2) = 7$.

The mean of an arithmetic progression is determined by the average of its first and last terms. Given that the mean of the 100 terms is 200, the arithmetic sum is $100 \times 200 = 20000$. Using the formula $S_n = \frac{n}{2}(2a + (n-1)d)$, where $n=100$ and $a=2$, we have $20000 = 50(4 + 99d)$, which simplifies to $4 + 99d = 400$, resulting in $d=4$.

The general term of the progression is defined as $x_i = 2 + (i-1)4 = 4i - 2$. Substituting this into the definition of $y_i$ gives $y_i = i(4i - 2 - i) = i(3i - 2) = 3i^2 - 2i$. Finding the mean of these terms requires evaluating the expression $\frac{1}{100} \sum_{i=1}^{100} (3i^2 - 2i)$. Using the summation identities $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{i=1}^n i = \frac{n(n+1)}{2}$ with $n=100$, we calculate $3 \left( \frac{100 \times 101 \times 201}{600} \right) - 2 \left( \frac{100 \times 101}{200} \right)$. This expression simplifies to $101 \times 100.5 - 101$, or $101 \times 99.5$, which yields 10049.50.

The series consists of 20 terms that can be organized into 10 pairs, where each pair comprises a positive term $2(2k)^2$ and a negative term $-(2k+1)^2$ for $k = 1, 2, \dots, 10$. Simplifying the expression $2(4k^2) - (4k^2 + 4k + 1)$ for each pair yields $4k^2 - 4k - 1$. The sum of these 10 pairs is determined by evaluating $4\sum_{k=1}^{10} k^2 - 4\sum_{k=1}^{10} k - \sum_{k=1}^{10} 1$. Applying the standard summation formulas, the calculation becomes $4\left(\frac{10 \times 11 \times 21}{6}\right) - 4\left(\frac{10 \times 11}{2}\right) - 10$. Solving these values results in $1540 - 220 - 10$, which equals 1310.

The coefficients of consecutive terms in the binomial expansion of $(1+x)^n$ are governed by the property $\frac{{}^nC_k}{{}^nC_{k-1}} = \frac{n-k+1}{k}$. When given the ratio of three consecutive coefficients as $1:5:20$, we can express these relationships as two separate ratios. The ratio of the first two terms is $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{5}{1}$, which simplifies to the equation $\frac{n-r+1}{r} = 5$, or $n = 6r-1$. Comparing the second and third terms gives the ratio $\frac{{}^nC_{r+1}}{{}^nC_r} = \frac{20}{5} = 4$, which leads to $\frac{n-r}{r+1} = 4$, simplifying to $n = 5r+4$.

Equating these two expressions for $n$, we find $6r-1 = 5r+4$, which yields $r = 5$. Substituting this back into either expression determines the power $n = 29$. Because the expansion starts with the 0th index, the fourth term in the expansion corresponds to the coefficient ${}^{29}C_3$. Calculating this value involves computing $\frac{29 \times 28 \times 27}{3 \times 2 \times 1}$, which results in 3654.

The key to solving this summation lies in the symmetry property of binomial coefficients, which states that ${}^{n}C_{r} = {}^{n}C_{n-r}$. By applying this identity to the second term, we can rewrite ${}^{23}C_{r}$ as ${}^{23}C_{23-r}$, transforming the expression into $\sum_{r=0}^{22} {}^{22}C_{r} \cdot {}^{23}C_{23-r}$. This form is a classic application of Vandermonde's Identity, which tells us that the sum of the products of binomial coefficients $\sum {}^{n}C_{k} \cdot {}^{m}C_{p-k}$ is equal to ${}^{n+m}C_{p}$. Here, we identify $n=22$, $m=23$, and the resulting index $p=23$. Therefore, the entire sum simplifies directly to ${}^{22+23}C_{23}$, which is equal to ${}^{45}C_{23}$.