Binomial Theorem – JEE Main MathematicsPractice Questions & PYQs

Expanding the expression inside the sigma notation allows us to leverage the binomial theorem to simplify the entire sum into a standard form. By observing that the term $\frac{10^{r+1}-1}{10^r}$ can be decomposed into $10 - \frac{1}{10^r}$, we can split the summation into two distinct parts: $10 \sum_{r=0}^{10} {}^{11}C_{r+1}$ and $\sum_{r=0}^{10} \frac{{}^{11}C_{r+1}}{10^r}$.

For the first part, evaluating $10 \sum_{r=0}^{10} {}^{11}C_{r+1}$ involves summing all binomial coefficients for $n=11$ excluding the first term, ${}^{11}C_0$. Since the sum of all binomial coefficients is $2^{11}$, this first component becomes $10(2^{11}-1)$.

For the second part, $\sum_{r=0}^{10} \frac{{}^{11}C_{r+1}}{10^r}$, we can adjust the index by setting $k=r+1$, transforming the expression into $10 \sum_{k=1}^{11} \frac{{}^{11}C_k}{10^k}$. This series corresponds to the binomial expansion of $(1+\frac{1}{10})^{11}$ minus the initial term, ${}^{11}C_0(\frac{1}{10})^0$. This evaluates to $10((\frac{11}{10})^{11}-1)$, which simplifies to $\frac{11^{11}-10^{11}}{10^{10}}$.

Combining these two results gives $10(2^{11}-1) - \frac{11^{11}-10^{11}}{10^{10}}$. When written over a common denominator of $10^{10}$, the expression becomes $\frac{10^{11}(2^{11}-1) - 11^{11} + 10^{11}}{10^{10}}$, which simplifies to $\frac{10^{11} \cdot 2^{11} - 10^{11} - 11^{11} + 10^{11}}{10^{10}}$. The $10^{11}$ terms cancel out, leaving $\frac{(10 \cdot 2)^{11} - 11^{11}}{10^{10}}$, or $\frac{20^{11} - 11^{11}}{10^{10}}$. Comparing this to the given structure $\frac{\alpha^{11}-11^{11}}{10^{10}}$, we identify that $\alpha$ must be 20.

The series for $b$ simplifies elegantly by applying the binomial identity $\sum_{k=0}^{n} {}^{n}C_{k} = 2^{n}$. By replacing the sum of the combinations in each term with the corresponding power of 2, the expression becomes $b = 1 + \frac{2^{1}}{1!} + \frac{2^{2}}{2!} + \frac{2^{3}}{3!} + \dots$, which matches the Taylor series expansion for the exponential function $e^{x}$ at $x=2$, confirming that $b = e^{2}$.

The value of $a$ is identified by examining the coefficients within the power series expansion of the function $\frac{e^{1+x}}{1+x}$. Through this functional analysis, the series converges to the value $a = \frac{e}{2}$. Having established these core values, the remaining task involves evaluating the required expression using these constants.

By substituting $b = e^{2}$ and $a = \frac{e}{2}$ into the formula $\frac{2b}{a^{2}}$, the expression becomes $\frac{2e^{2}}{(e/2)^{2}}$. Squaring the denominator gives $\frac{e^{2}}{4}$, and simplifying the fraction results in $2e^{2} \cdot \frac{4}{e^{2}}$, which yields 8.

The sequence $1, 4, 8, 13, 19, 26, \ldots$ exhibits a pattern in its first differences of $3, 4, 5, 6, 7, \ldots$, which signifies that the general term follows a quadratic progression. By setting the general term $t_n = an^2 + bn + c$ and solving for the coefficients using the initial terms, we determine that $t_n = \frac{n^2+3n-2}{2}$. Since $\alpha$ is defined as the sum of these squared terms, we have $\alpha = \sum_{n=1}^{10} \left(\frac{n^2+3n-2}{2}\right)^2$, which can be rearranged to $4\alpha = \sum_{n=1}^{10} (n^2+3n-2)^2$.

Expanding the expression $(n^2+3n-2)^2$ yields $n^4 + 6n^3 + 5n^2 - 12n + 4$. Subtracting $\beta = \sum_{n=1}^{10} n^4$ from this expansion eliminates the $n^4$ term, simplifying the expression to $4\alpha - \beta = \sum_{n=1}^{10} (6n^3 + 5n^2 - 12n + 4)$. Applying standard summation formulas for $n=10$, where the sum of cubes $\sum n^3 = 3025$, the sum of squares $\sum n^2 = 385$, and the sum of integers $\sum n = 55$, we calculate the total as $6(3025) + 5(385) - 12(55) + 40$. This computation results in $18150 + 1925 - 660 + 40$, which equals $19455$. Solving the final equation $55k + 40 = 19455$ gives $55k = 19415$, confirming that $k = 353$.

Identifying common terms within two arithmetic progressions requires finding the smallest value shared by both series and utilizing the least common multiple of their individual common differences. For the first sequence, the 25th term is calculated as $4 + (25-1) \times 5 = 124$, and for the second sequence, the 37th term is $3 + (37-1) \times 3 = 111$. Because all shared numbers must reside within both ranges, our search is limited by the smaller upper bound of 111.

The common differences for the two series are 5 and 3, meaning the overlapping values will form a new progression with a common difference of $\text{lcm}(5, 3) = 15$. Starting with the first intersection at 9, we generate subsequent terms by repeatedly adding 15, resulting in the set $9, 24, 39, 54, 69, 84, 99$. The next term, 114, falls outside the range of 111, confirming there are exactly 7 such terms.

The binomial expansion of $(x+y)^n$ generates successive terms whose ratios effectively cancel out common variables, allowing us to isolate $n, x,$ and $y$. By identifying the given terms as ${}^nC_1 x^{n-1} y = 135$, ${}^nC_2 x^{n-2} y^2 = 30$, and ${}^nC_3 x^{n-3} y^3 = \frac{10}{3}$, we can take the ratios of consecutive terms to eliminate variables and solve for $n$.

Dividing the second term by the third yields $\frac{2}{n-1} \cdot \frac{x}{y} = \frac{135}{30} = \frac{9}{2}$, and dividing the third term by the fourth yields $\frac{3}{n-2} \cdot \frac{x}{y} = \frac{30}{10/3} = 9$. Solving these for $\frac{x}{y}$ provides $\frac{x}{y} = \frac{9(n-1)}{4}$ and $\frac{x}{y} = 3(n-2)$, which when equated result in $3(n-1) = 4(n-2)$, giving $n = 5$. With $n = 5$ confirmed, the ratio $\frac{x}{y} = 3(5-2) = 9$ implies $x = 9y$. Substituting $x = 9y$ into the first term equation ${}^5C_1 x^4 y = 135$ results in $5 x^4 (\frac{x}{9}) = 135$, or $x^5 = 243$, leading to $x = 3$ and $y = \frac{1}{3}$. Evaluating the expression $6(n^3 + x^2 + y)$ with these values yields $6(125 + 9 + \frac{1}{3}) = 806$.

The general term for the expansion of $(a+b)^n$ is defined by the formula $t_{r+1} = {}^{n}C_r a^{n-r} b^r$. For the binomial expansion $\left( \frac{1}{3} x^{1/3} + \frac{1}{2 x^{2/3}} \right)^{18}$, the seventh term occurs when $r=6$ and the thirteenth term occurs when $r=12$.

For the seventh term, $m = {}^{18}C_6 \left( \frac{1}{3} \right)^{12} \left( \frac{1}{2} \right)^6$, while for the thirteenth term, $n = {}^{18}C_{12} \left( \frac{1}{3} \right)^6 \left( \frac{1}{2} \right)^{12}$. Because the combination terms are identical, with ${}^{18}C_6 = {}^{18}C_{12}$, the ratio $\frac{n}{m}$ simplifies to the expression $\frac{3^{-6} \cdot 2^{-12}}{3^{-12} \cdot 2^{-6}}$. Simplifying the powers of $3$ and $2$ yields $3^6 \cdot 2^{-6}$, which is $\left( \frac{3}{2} \right)^6$. Taking the cube root of this ratio results in $\left( \frac{3}{2} \right)^2$, which equals $\frac{9}{4}$.

Simplifying the expression $\left(1 + \frac{1}{x}\right)^6 (1 + x^2)^7 (1 - x^3)^8$ as $\frac{(1 + x)^6 (1 + x^2)^7 (1 - x^3)^8}{x^6}$ reveals that the coefficient of $x^{30}$ in the original expression corresponds to the coefficient of $x^{36}$ in the expanded numerator $(1 + x)^6 (1 + x^2)^7 (1 - x^3)^8$.

Using the multinomial expansion, the general term is given by ${}^6C_{r_1} {}^7C_{r_2} {}^8C_{r_3} (-1)^{r_3} x^{r_1 + 2r_2 + 3r_3}$. We must find all non-negative integer solutions for $r_1, r_2, r_3$ that satisfy the equation $r_1 + 2r_2 + 3r_3 = 36$ while keeping $0 \le r_1 \le 6$, $0 \le r_2 \le 7$, and $0 \le r_3 \le 8$.

Considering the possible values for $r_3$, we investigate the scenarios for $r_3 = 8$, $r_3 = 7$, and $r_3 = 6$. When $r_3 = 8$, we require $r_1 + 2r_2 = 12$, which provides combinations of $(r_1, r_2, r_3)$ as $(0, 6, 8)$, $(2, 5, 8)$, $(4, 4, 8)$, and $(6, 3, 8)$ with coefficients 7, 315, 525, and 35. Setting $r_3 = 7$ leads to $r_1 + 2r_2 = 15$, yielding combinations $(1, 7, 7)$, $(3, 6, 7)$, and $(5, 5, 7)$ which contribute -48, -1120, and -1008 due to the $(-1)^{r_3}$ factor. Finally, when $r_3 = 6$, we need $r_1 + 2r_2 = 18$, providing combinations $(4, 7, 6)$ and $(6, 6, 6)$ with coefficients 420 and 196.

Summing these individual terms, 882 - 2176 + 616, yields a total value for $\alpha$ of -678. Therefore, taking the absolute value gives the final result of 678.

The coefficient of $x^{70}$ in the general term $x^k(1+x)^{100-k}$ corresponds to the coefficient of $x^{70-k}$ in the expansion of $(1+x)^{100-k}$, which is given by the binomial coefficient ${}^{100-k}C_{70-k}$. Expanding this for the given range, where $k$ ranges from $2$ to $54$, yields the summation ${}^{98}C_{68} + {}^{97}C_{67} + \dots + {}^{46}C_{16}$. By applying the symmetry property ${}^nC_r = {}^nC_{n-r}$, each term simplifies to ${}^{100-k}C_{30}$, resulting in the series ${}^{98}C_{30} + {}^{97}C_{30} + \dots + {}^{46}C_{30}$.

Using the hockey-stick identity, which states that $\sum_{i=r}^n {}^iC_r = {}^{n+1}C_{r+1}$, we can evaluate this sum by calculating the difference between the complete sequence from $i=30$ to $98$ and the missing sequence from $i=30$ to $45$. This arithmetic results in the expression ${}^{99}C_{31} - {}^{46}C_{31}$. Given the symmetry property ${}^nC_r = {}^nC_{n-r}$, the variables $p$ and $q$ can alternatively take values $p \in \{31, 68\}$ and $q \in \{31, 15\}$. Considering all possible combinations, the sums $p+q$ can be $62, 46, 99,$ or $83$, identifying $83$ as the valid value from the provided choices.

The given series represents an infinite arithmetico-geometric progression that can be resolved by splitting it into two separate components. The constant portion $3 + \frac{3}{4} + \frac{3}{16} + \dots$ acts as a geometric progression summing to $\frac{3}{1 - 1/4} = 4$. Simultaneously, the terms involving $p$ follow the pattern $p(\frac{1}{4} + \frac{2}{16} + \frac{3}{64} + \dots)$, which corresponds to the standard sum $p \cdot \sum_{n=1}^{\infty } n r^n = p \cdot \frac{r}{(1-r)^2}$ with $r = \frac{1}{4}$, evaluating to $p \cdot \frac{1/4}{(3/4)^2} = \frac{4p}{9}$. Setting the sum of these two components equal to 8 results in the equation $4 + \frac{4p}{9} = 8$, which simplifies to $\frac{4p}{9} = 4$ and confirms that $p = 9$.

For any sequence of three terms to exist in an arithmetic progression, the middle term must be the arithmetic mean of the other two, which implies that $2 \log_e b = \log_e a + \log_e c$. By utilizing logarithmic properties, this expression simplifies to $\log_e b^2 = \log_e (ac)$, leading to the fundamental constraint $b^2 = ac$.

Similarly, when the terms $\log_e \left( \frac{a}{2b} \right)$, $\log_e \left( \frac{2b}{3c} \right)$, and $\log_e \left( \frac{3c}{a} \right)$ are in an arithmetic progression, their sum must equal twice the middle term, written as $2 \log_e \left( \frac{2b}{3c} \right) = \log_e \left( \frac{a}{2b} \right) + \log_e \left( \frac{3c}{a} \right)$. Combining the logarithms on the right side produces $\log_e \left( \frac{a}{2b} \cdot \frac{3c}{a} \right)$, which simplifies to $\log_e \left( \frac{3c}{2b} \right)$. Equating the arguments of the logarithms leads to $\left( \frac{2b}{3c} \right)^2 = \frac{3c}{2b}$, or $\left( \frac{2b}{3c} \right)^3 = 1$, which confirms that $\frac{b}{c} = \frac{3}{2}$.

With the ratio $\frac{b}{c} = \frac{3}{2}$, we have $c = \frac{2b}{3}$. Substituting this into our first identity $b^2 = ac$ yields $b^2 = a \left( \frac{2b}{3} \right)$, which simplifies to $\frac{a}{b} = \frac{3}{2}$. Expressing both $a$ and $c$ in terms of $b$ gives $a = \frac{3}{2}b$ and $c = \frac{2}{3}b$. The final ratio $a : b : c$ is therefore $\frac{3}{2} : 1 : \frac{2}{3}$, which results in $9 : 6 : 4$.

The total sum of a geometric progression is calculated using the formula $S = \frac{a(r^n - 1)}{r-1}$, where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. For a progression with 64 terms, the odd-positioned terms form their own geometric progression with 32 terms, the first term $a$, and a common ratio of $r^2$. The sum of these 32 odd-positioned terms is expressed as $\frac{a((r^2)^{32}-1)}{r^2-1}$, which simplifies to $\frac{a(r^{64}-1)}{r^2-1}$.

Equating the total sum of all 64 terms to 7 times the sum of these odd terms yields the equality $\frac{a(r^{64}-1)}{r-1} = \frac{7a(r^{64}-1)}{r^2-1}$. Dividing both sides by the common factor $a(r^{64}-1)$, the relationship simplifies to $\frac{1}{r-1} = \frac{7}{r^2-1}$. Factoring $r^2-1$ as $(r-1)(r+1)$ allows the equation to become $\frac{1}{r-1} = \frac{7}{(r-1)(r+1)}$, which implies $1 = \frac{7}{r+1}$. Solving this final linear equation gives $r+1 = 7$, resulting in a common ratio of 6.

The condition that each term is the arithmetic mean of its two successors defines the relation $2 a_n = a_{n+1} + a_{n+2}$. Substituting the geometric progression term $a_n = a_1 r^{n-1}$ into this relationship yields $2 a_1 r^{n-1} = a_1 r^n + a_1 r^{n+1}$, which simplifies to $r^2 + r - 2 = 0$ after dividing by $a_1 r^{n-1}$. Since the term $a_2 \neq a_1$ implies $r \neq 1$, we must have $r = -2$.

Calculating $S_{20} - S_{18}$ isolates the 19th and 20th terms as $a_1 r^{18} + a_1 r^{19}$, which simplifies to $a_1 r^{18}(1 + r)$. Inserting $a_1 = \frac{1}{8}$ and $r = -2$ gives $\frac{1}{8} \cdot (-2)^{18} \cdot (1 - 2)$. Since $(-2)^{18}$ is equal to $2^{18}$, the calculation reduces to $\frac{1}{8} \cdot 2^{18} \cdot (-1)$, resulting in $-2^{15}$.

Finding a term from the end of an arithmetic progression is most easily handled by reversing the sequence. In the original progression, the common difference is negative, so reversing the sequence makes the final term, $-129 \frac{1}{4}$, the new starting point and changes the common difference to positive $\frac{3}{4}$. Using the general formula for the $n^{th}$ term, $T_n = a + (n - 1)d$, where $n = 20$, $a = -129 \frac{1}{4}$, and $d = \frac{3}{4}$, we can determine the value directly. This calculation, $T_{20} = -129 \frac{1}{4} + 19 \left(\frac{3}{4}\right)$, simplifies to $-115$.

Finding the term independent of $x$ in a binomial expansion relies on the general term formula, which expresses any term as $T_{r+1} = {}^{10}C_r (\sqrt{a} x^2)^{10-r} \left( \frac{1}{2 x^3} \right)^r$ Separating the variables and the constants allows us to combine the powers of $x$ into $x^{20-2r} \cdot x^{-3r}$, which simplifies to $x^{20-5r}$. An independent term contains no variable, so we set the exponent to zero, giving the equation $20 - 5r = 0$, which identifies $r = 4$.

Substituting this value into the expression produces the coefficient ${}^{10}C_4 (\sqrt{a})^6 \left( \frac{1}{2} \right)^4 = 105$ Simplifying this leads to $210 \cdot a^3 \cdot \frac{1}{16} = 105$, which further reduces to $a^3 = 8$. Given $a = 2$, we find that $a^2 = 4$.

The most effective way to determine the remainder of a large exponential expression is to simplify the base using modular arithmetic, which allows us to replace the base with its remainder when divided by the divisor. Dividing 428 by 21 shows that $428 = 21 \times 20 + 8$, which means that $428 \equiv 8 \pm od{21}$. This reduces the original expression $428^{2024}$ to evaluating $8^{2024} \pm od{21}$.

We can handle this exponent by observing that $8^2 = 64$. Because $64 = 3 \times 21 + 1$, it follows that $8^2$ leaves a remainder of 1 when divided by 21. By raising this relationship to the power of 1012, we obtain $(8^2)^{1012} = 8^{2024} \equiv 1^{1012} \pm od{21}$, which simplifies to 1. Therefore, the final remainder is 1.

The binomial coefficients of the expansion $(1+x)^n$ corresponding to $x^4$, $x^5$, and $x^6$ are represented by ${}^n C_4$, ${}^n C_5$, and ${}^n C_6$, respectively. For these three terms to exist in an arithmetic progression, the middle term must satisfy the condition that twice its value equals the sum of the adjacent terms, which is expressed as $2 \cdot {}^n C_5 = {}^n C_4 + {}^n C_6$.

To solve for $n$, divide the entire equation by ${}^n C_5$ to obtain $2 = \frac{{}^n C_4}{{}^n C_5} + \frac{{}^n C_6}{{}^n C_5}$. By applying the ratio property $\frac{{}^n C_r}{{}^n C_{r-1}} = \frac{n-r+1}{r}$, the specific ratios become $\frac{{}^n C_4}{{}^n C_5} = \frac{5}{n-4}$ and $\frac{{}^n C_6}{{}^n C_5} = \frac{n-5}{6}$. Substituting these expressions results in the equation $2 = \frac{5}{n-4} + \frac{n-5}{6}$.

Multiplying the entire equation by $6(n-4)$ to eliminate the denominators gives $12(n-4) = 30 + (n-5)(n-4)$. Expanding the right side leads to $12n - 48 = 30 + n^2 - 9n + 20$, which simplifies to the quadratic equation $n^2 - 21n + 98 = 0$. Factoring this quadratic reveals $(n-14)(n-7) = 0$, providing two possible values for $n$: $7$ and $14$. Identifying the maximum value among these possibilities confirms that $14$ is the result.

The general term in the binomial expansion of $(a+b)^n$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$. For the expansion $(x^{2/3}+\frac{1}{2} x^{-2/5})^9$, the general term simplifies as
$T_{r+1} = {}^{9}C_{r} \left(\frac{1}{2}\right)^{r} x^{6 - \frac{16r}{15}}$
Equating the exponent $6 - \frac{16r}{15}$ to $\frac{2}{3}$ identifies $r=5$, which yields a coefficient of ${}^{9}C_{5} (\frac{1}{2})^{5} = \frac{63}{16}$. Setting the exponent $6 - \frac{16r}{15}$ equal to $-\frac{2}{5}$ yields $r=6$, providing a coefficient of ${}^{9}C_{6} (\frac{1}{2})^{6} = \frac{21}{16}$. The sum of these two coefficients is
$\frac{63}{16} + \frac{21}{16} = \frac{21}{4}$

In the binomial expansion of $(2^{\frac{1}{5}} + 5^{\frac{1}{3}})^{15}$, a term is rational if and only if the exponents of the bases are integers. Given the general term formula $T_{r+1} = {}^{15}C_r (2^{\frac{1}{5}})^{15-r} (5^{\frac{1}{3}})^r$, the expression simplifies to ${}^{15}C_r \cdot 2^{\frac{15-r}{5}} \cdot 5^{\frac{r}{3}}$, where $r$ must be a multiple of both $5$ and $3$ to ensure the powers are integers. These conditions restrict $r$ to be a multiple of $15$ within the range $0 \le r \le 15$, identifying $r=0$ and $r=15$ as the only valid terms. Calculating these results in ${}^{15}C_0 \cdot 2^3 \cdot 5^0 = 8$ and ${}^{15}C_{15} \cdot 2^0 \cdot 5^5 = 3125$, which produce a final sum of $8 + 3125 = 3133$.

The relationship $a_6 = a + 5d = 2$ allows the first term of the progression to be expressed as $a = 2 - 5d$. Substituting this into the product of the first, fourth, and fifth terms, $a(a + 3d)(a + 4d)$, results in the function $f(d) = (2 - 5d)(2 - 2d)(2 - d)$. Expanding this expression yields the cubic polynomial $f(d) = -10d^3 + 34d^2 - 32d + 8$. To identify the common difference that maximizes this product, we determine the derivative $f'(d) = -30d^2 + 68d - 32$ and set it equal to zero. Factoring the resulting quadratic equation $-2(5d - 8)(3d - 2) = 0$ provides the critical points $d = \frac{8}{5}$ and $d = \frac{2}{3}$, with $d = \frac{8}{5}$ yielding the maximum value.

Identifying the constant term in any binomial expansion involves isolating the variable $x$ within the general term formula to find the value of $r$ that yields an exponent of zero. For the expansion $\left( \frac{\sqrt[5]{3}}{x} + \frac{2x}{\sqrt[3]{5}} \right)^{12}$, the general term is defined as $T_{r+1} = {}^{12}C_r \left( \frac{3^{1/5}}{x} \right)^{12-r} \left( \frac{2x}{5^{1/3}} \right)^r$, which simplifies by grouping into a coefficient and a variable part $x^{r - (12-r)} = x^{2r-12}$. Setting the exponent $2r-12$ to zero reveals that the constant term occurs at $r = 6$.

Substituting $r = 6$ into the general expression yields $T_7 = {}^{12}C_6 \frac{(3^{1/5})^{6} (2x)^6}{x^6 (5^{1/3})^6}$. Simplifying this expression results in $T_7 = \frac{924 \times 3^{6/5} \times 2^6}{25}$. By expressing $3^{6/5}$ as $3 \times 3^{1/5}$ and rewriting the constant to match the form $\alpha \times 2^8 \times \sqrt[5]{3}$, we process the constants $924 \times 3 \times 2^6$. Dividing this product by $2^2$ to elevate $2^6$ to $2^8$ gives $693 \times 2^8 \times 3^{1/5} / 25$. Therefore, the value of $\alpha$ is $\frac{693}{25}$, which means $25 \alpha = 693$.