Binomial Theorem – JEE Main MathematicsPractice Questions & PYQs

The Binomial Theorem determines specific coefficients using the general term $T_{r+1} = \binom{n}{r} A^{n-r} B^r$. For the expression $(ax^2 + \frac{1}{2bx})^{11}$, the general term is $\binom{11}{r} a^{11-r} x^{22-2r} (2b)^{-r} x^{-r}$, which simplifies to $\binom{11}{r} a^{11-r} (2b)^{-r} x^{22-3r}$. Setting the exponent $22 - 3r = 7$ yields $r = 5$, making the coefficient for $x^7$ equal to $\binom{11}{5} a^6 \frac{1}{32b^5}$.

Expanding $(ax - \frac{1}{3bx^2})^{11}$ follows the same process, where the general term $\binom{11}{r} (ax)^{11-r} (-\frac{1}{3bx^2})^r$ results in a variable component of $x^{11-3r}$. Solving $11 - 3r = -7$ gives $r = 6$, providing a coefficient of $\binom{11}{6} a^5 \frac{1}{729b^6}$. Equating the two derived expressions, $\binom{11}{5} \frac{a^6}{32b^5} = \binom{11}{6} \frac{a^5}{729b^6}$, causes the binomial combinations to cancel out because $\binom{11}{5} = \binom{11}{6}$. Simplifying the remaining powers of $a$ and $b$ results in $\frac{a}{32} = \frac{1}{729b}$, which rearranges to $729 ab = 32$.

The binomial coefficient $a_r$ denotes the coefficient of $x^{10-r}$ in the expansion of $(1+x)^{10}$. Due to the property ${}^{n}C_k = {}^{n}C_{n-k}$, $a_r$ equals ${}^{10}C_{10-r}$, which simplifies to ${}^{10}C_r$. The ratio $\frac{a_r}{a_{r-1}}$ thus becomes $\frac{{}^{10}C_r}{{}^{10}C_{r-1}}$, and applying the standard reduction formula $\frac{{}^{n}C_r}{{}^{n}C_{r-1}} = \frac{n-r+1}{r}$ results in $\frac{10-r+1}{r}$, or $\frac{11-r}{r}$.

Evaluating the target summation involves substituting this ratio into the original expression:
$\sum_{r=1}^{10} r^3 \left( \frac{11-r}{r} \right)^2 = \sum_{r=1}^{10} r^3 \frac{(11-r)^2}{r^2} = \sum_{r=1}^{10} r(11-r)^2$
Expanding this expression yields:
$\sum_{r=1}^{10} (121r - 22r^2 + r^3) = 121 \sum_{r=1}^{10} r - 22 \sum_{r=1}^{10} r^2 + \sum_{r=1}^{10} r^3$
Using the standard sum formulas for $n=10$, where $\sum r = 55$, $\sum r^2 = 385$, and $\sum r^3 = 3025$, the calculation becomes $121(55) - 22(385) + 3025$. This evaluates to $6655 - 8470 + 3025$, resulting in 1210.

Calculating the sum of an arithmetic progression with $n=12$ terms involves the formula $\frac{n}{2} [2a + (n-1)d]$. For the $k$-th progression, where the first term is $k$ and the common difference follows the sequence $1, 3, 5, \dots, 19$ defined by $d = 2k-1$, the sum $s_k$ simplifies to $6(2k + 11(2k-1))$, which reduces further to $144k - 66$. To find the total sum across all 10 progressions, we evaluate the expression $\sum_{k=1}^{10} (144k - 66)$, which separates into $144 \sum_{k=1}^{10} k - 660$. Substituting the sum of the first 10 integers, which is $\frac{10 \times 11}{2} = 55$, we calculate $144 \times 55 - 660$, ultimately yielding $7920 - 660 = 7260$.

The multinomial theorem allows the identification of individual terms in an expansion of the form $(a + b + c)^n$ through the general term expression $\frac{n!}{n_1! n_2! n_3!} a^{n_1} b^{n_2} c^{n_3}$, where the sum of indices $n_1 + n_2 + n_3 = n$ must be satisfied. For the expansion $\left(2x + \frac{1}{x^7} + 3x^2\right)^5$, the variable part is determined by the product $x^{n_1} \cdot (x^{-7})^{n_2} \cdot (x^2)^{n_3}$, which simplifies to $x^{n_1 - 7n_2 + 2n_3}$. Because the goal is to find the constant term, the exponent must vanish, setting the condition $n_1 - 7n_2 + 2n_3 = 0$.

Substituting $n_1 = 5 - n_2 - n_3$ into this exponent equation leads to $5 - n_2 - n_3 + 2n_3 = 7n_2$, which simplifies to $5 + n_3 = 8n_2$. Since $n_1, n_2,$ and $n_3$ are non-negative integers that must sum to 5, testing possible values confirms that $n_2 = 1$ is the only valid solution, which forces $n_3 = 3$ and consequently $n_1 = 1$. Substituting these specific values into the coefficient portion $\frac{5!}{n_1! n_2! n_3!} (2)^{n_1} (1)^{n_2} (3)^{n_3}$ results in $\frac{120}{1! \cdot 1! \cdot 3!} \cdot 2^1 \cdot 1^1 \cdot 3^3$, which calculates to $20 \cdot 2 \cdot 27 = 1080$.

The binomial theorem provides a straightforward method to isolate the coefficient of any specific power of $x$ by utilizing the general term formula, $T_{r+1} = ^nC_r a^{n-r}b^r$. For the expansion of $(2x^2 + \frac{1}{2x})^{11}$, the general term is given by $T_{r+1} = ^{11}C_r (2x^2)^{11-r} (\frac{1}{2x})^r$. Simplifying this allows us to group the numerical constants and the variable components, resulting in $T_{r+1} = ^{11}C_r \cdot 2^{11-2r} \cdot x^{22-3r}$.

To find the coefficient of $x^{10}$, we set the exponent $22-3r = 10$, which implies $3r = 12$ or $r = 4$. Substituting $r=4$ into the simplified expression gives the coefficient $^{11}C_4 \cdot 2^{11-8} = ^{11}C_4 \cdot 2^3$. Similarly, to find the coefficient of $x^7$, we set $22-3r = 7$, which leads to $3r = 15$ or $r = 5$, yielding the coefficient $^{11}C_5 \cdot 2^{11-10} = ^{11}C_5 \cdot 2^1$.

The absolute difference between these coefficients is $| (^{11}C_4 \cdot 2^3) - (^{11}C_5 \cdot 2^1) |$. Calculating the binomial values, $^{11}C_4$ equals 330 and $^{11}C_5$ equals 462. Performing the arithmetic gives $330 \cdot 8 - 462 \cdot 2$, which is $2640 - 924$, resulting in 1716. Expressing this value in the form $n^3-n$ shows it is equivalent to $12^3 - 12$.

This series is a geometric progression consisting of 501 terms, where the first term is $(1+x)^{500}$ and the common ratio is $\frac{x}{1+x}$. By applying the formula for the sum of a finite geometric series, $S_n = a\frac{1-r^n}{1-r}$, and simplifying the resulting algebraic expression, the entire series reduces to the much simpler form $(1+x)^{501} - x^{501}$.

Extracting the coefficient of $x^{301}$ requires looking at the binomial expansion of $(1+x)^{501}$, as the term $-x^{501}$ does not contribute to the power of $x^{301}$. The coefficient is then represented by the binomial coefficient ${}^{501}C_{301}$. Utilizing the symmetry property of combinations, where ${}^{n}C_{r} = {}^{n}C_{n-r}$, this expression becomes ${}^{501}C_{501-301}$, which simplifies to ${}^{501}C_{200}$.

For the 9th term to be the greatest in the binomial expansion of $(3+6x)^n$, it must be simultaneously greater than or equal to both the 8th and 10th terms. We can determine these conditions by examining the ratios $\frac{T_9}{T_8} \ge 1$ and $\frac{T_{10}}{T_9} \le 1$. Utilizing the general term formula $T_{r+1} = {n \choose r} 3^{n-r} (6x)^r$ and setting $x = \frac{3}{2}$, the ratio $\frac{T_9}{T_8}$ becomes $\frac{{n \choose 8} 3^{n-8} (6 \cdot 3/2)^8}{{n \choose 7} 3^{n-7} (6 \cdot 3/2)^7}$, which simplifies to $\frac{n-7}{8} \cdot \frac{6 \cdot 3/2}{3} = \frac{3(n-7)}{8}$. Requiring this to be at least $1$ leads to $3n - 21 \ge 8$, or $n \ge \frac{29}{3}$, meaning $n \ge 9.66$. Similarly, the ratio $\frac{T_{10}}{T_9}$ simplifies to $\frac{n-8}{9} \cdot \frac{6 \cdot 3/2}{3} = \frac{n-8}{3}$, and requiring this to be at most $1$ leads to $n-8 \le 3$, or $n \le 11$. The integers satisfying these bounds are $10$ and $11$, making the least value $n_0 = 10$.

With $n=10$, we calculate the ratio $k$ of the coefficient of $x^6$ to the coefficient of $x^3$ in $(3+6x)^{10}$. The coefficient of $x^r$ is ${10 \choose r} 3^{10-r} 6^r$. Thus, the coefficient of $x^6$ is ${10 \choose 6} 3^4 6^6$ and the coefficient of $x^3$ is ${10 \choose 3} 3^7 6^3$. The ratio $k$ is defined as $\frac{{10 \choose 6} 3^4 6^6}{{10 \choose 3} 3^7 6^3}$. Simplifying the binomial coefficients, we have $\frac{210}{120} = \frac{7}{4}$. The powers of $3$ and $6$ simplify as $\frac{3^4}{3^7} \cdot \frac{6^6}{6^3} = \frac{1}{3^3} \cdot 6^3 = \frac{216}{27} = 8$. Combining these, we find $k = \frac{7}{4} \cdot 8 = 14$. The final result is $k + n_0 = 14 + 10 = 24$.

Modular arithmetic allows us to simplify expressions by examining the remainders of the bases when divided by the specified divisors. For the expression $(22)^{2022} + (2022)^{22}$ divided by 3, we note that $22 \equiv 1 \pm od 3$ and $2022 \equiv 0 \pm od 3$. This simplifies to $1^{2022} + 0^{22}$, which gives a remainder of $\alpha = 1$.

To find the remainder when dividing by 7, we observe that $22 \equiv 1 \pm od 7$ since 22 is 1 more than 21. Similarly, $2022 \equiv -1 \pm od 7$ because 2023 is a multiple of 7. Substituting these into the expression yields $1^{2022} + (-1)^{22}$, which simplifies to $1 + 1 = 2$, identifying $\beta = 2$. Combining these results, the final calculation is $\alpha^2 + \beta^2 = 1^2 + 2^2 = 5$.

The binomial expansion theorem provides the general term for the expansion of $(a+b)^n$ as $T_{r+1} = {}^{n}C_r a^{n-r} b^r$. For the expression $\left(2 x^3 + \frac{3}{x^k}\right)^{12}$, the general term is given by $T_{r+1} = {}^{12}C_r (2x^3)^r (3x^{-k})^{12-r}$, which simplifies to $T_{r+1} = {}^{12}C_r \cdot 2^r \cdot 3^{12-r} \cdot x^{3r - k(12-r)}$. To isolate the constant term, the exponent of $x$ must be zero, leading to the condition $3r - k(12-r) = 0$, or $k = \frac{3r}{12-r}$.

Since $k$ must be a positive integer and $r$ can take any integer value from 0 to 11, we evaluate the possible pairs of $(r, k)$ that result in an integer $k$. Solving for $r \in \{0, 1, \dots, 11\}$ yields pairs $(r, k)$ of $(3, 1)$, $(6, 3)$, $(8, 6)$, $(9, 9)$, and $(10, 15)$. We then identify which of these cases results in a constant term of the form $2^8 \cdot I$, where $I$ is an odd integer. For $r=3$, the constant term is ${}^{12}C_3 \cdot 2^3 \cdot 3^9 = 220 \cdot 2^3 \cdot 3^9 = 55 \cdot 2^5 \cdot 3^9$, which does not contain the required power of 2. For $r=6$, the term is ${}^{12}C_6 \cdot 2^6 \cdot 3^6 = 924 \cdot 2^6 \cdot 3^6 = (231 \cdot 4) \cdot 2^6 \cdot 3^6 = 231 \cdot 2^8 \cdot 3^6$, where 231 is odd, making this a valid solution. For $r=8$, the term is ${}^{12}C_8 \cdot 2^8 \cdot 3^4 = 495 \cdot 2^8 \cdot 3^4$, where 495 is odd, which also qualifies as a valid solution. For $r=9$, the calculation ${}^{12}C_9 \cdot 2^9 \cdot 3^3 = 220 \cdot 2^9 \cdot 3^3 = 55 \cdot 2^{11} \cdot 3^3$ does not meet the condition, and similarly for $r=10$, we obtain ${}^{12}C_{10} \cdot 2^{10} \cdot 3^2 = 66 \cdot 2^{10} \cdot 3^2 = 33 \cdot 2^{11} \cdot 3^2$, which also fails. Consequently, there are two values of $k$ that satisfy the condition.

The series $1+3+3^2+3^3+\cdot s+3^{2021}$ forms a finite geometric progression with 2022 terms, where the first term is 1 and the common ratio is 3. Applying the standard formula for the sum of a geometric progression, the total is equivalent to $\frac{3^{2022}-1}{3-1}$, which simplifies to $\frac{3^{2022}-1}{2}$.

Determining the remainder upon division by 50 involves evaluating the expression modulo 50, which relates to calculating $3^{2022}-1$ modulo 100. By noting that $3^{2022} = (3^2)^{1011} = (10-1)^{1011}$, one can utilize the binomial expansion to evaluate the power. This expansion yields a sum of terms involving powers of 10, specifically $(10-1)^{1011} = \sum_{k=0}^{1011} \binom{1011}{k} 10^k (-1)^{1011-k}$. All terms with $k \ge 2$ are multiples of 100, leaving the two final terms as $1011 \cdot 10^1 \cdot (-1)^{1010} + 1 \cdot (-1)^{1011}$, which calculates to $10110 - 1 = 10109$.

Because $3^{2022}$ is equivalent to 10109 modulo 100, the numerator $3^{2022}-1$ is equivalent to 10108 modulo 100. Substituting this into the sum formula gives $\frac{10108}{2}$ modulo 50, resulting in $5054 \pm od{50}$. Since $5054 = 50 \times 101 + 4$, the final remainder is 4.

Summing a factorial-based sequence often requires identifying a pattern where terms cancel out, known as a telescoping sum. We can rewrite the general term $(r^2+1)(r!)$ by splitting the $r^2$ term to group it with the factorial. Specifically, expressing the term as $r(r+1)! - (r-1)r!$ effectively decomposes the expression into a difference of the form $f(r) - f(r-1)$, which simplifies the entire summation significantly.

Applying this to the summation $\sum_{r=1}^{20} [r(r+1)! - (r-1)r!]$, we can observe the pattern by listing the first few terms. For $r=1$, we have $1(2!) - 0(1!)$; for $r=2$, we get $2(3!) - 1(2!)$; for $r=3$, we obtain $3(4!) - 2(3!)$, and this sequence continues until $r=20$, which yields $20(21!) - 19(20!)$. Notice how the positive part of each term cancels exactly with the negative part of the succeeding term. After all intermediate values cancel out, only the positive part of the final term and the negative part of the first term remain, resulting in $20(21!)$. To express this result in the factorial notation used in the options, we recognize that $22! - 2(21!) = 22 \times 21! - 2 \times 21!$, which simplifies to $20 \times 21!$, matching the final value of the telescoping series.

The total sum $\sum_{\substack{i,j=0 \\ i\neq j}}^{n} {}^n C_i {}^n C_j$ is evaluated by considering the sum over all possible combinations of $i$ and $j$ ranging from $0$ to $n$ and then removing the terms where $i = j$. The total sum over all $i$ and $j$ is equivalent to the product of two independent summations, $(\sum_{i=0}^n {}^n C_i) \cdot (\sum_{j=0}^n {}^n C_j)$, which equals $2^n \cdot 2^n = 2^{2n}$. To satisfy the restriction $i \neq j$, we must subtract the diagonal terms $\sum_{i=j=0}^n ({}^n C_i)^2$. By Vandermonde's Identity, this sum of squares is equal to ${}^{2n} C_n$. Subtracting this from the total result yields the final expression $2^{2n} - {}^{2n} C_n$.

The binomial expansion of $\left(x^n+\frac{2}{x^5}\right)^7$ contains a general term given by $T_{r+1} = {}^7 C_r (x^n)^{7-r} (2x^{-5})^r$, which simplifies to ${}^7 C_r \cdot 2^r \cdot x^{7n - nr - 5r}$. A term has a positive power of $x$ when the exponent $7n - nr - 5r$ is greater than zero, which leads to the condition $r(n+5) < 7n$, or $r < \frac{7n}{n+5}$. Since $r$ must be an integer, we determine which terms possess positive powers by calculating the cumulative sum of the coefficients ${}^7 C_r \cdot 2^r$ for consecutive values of $r$.

Starting from $r=0$, the coefficient is ${}^7 C_0 \cdot 2^0 = 1$. For $r=1$, the sum is $1 + ({}^7 C_1 \cdot 2^1) = 1 + 14 = 15$. For $r=2$, the sum is $15 + ({}^7 C_2 \cdot 2^2) = 15 + 84 = 99$. For $r=3$, the sum becomes $99 + ({}^7 C_3 \cdot 2^3) = 99 + 280 = 379$. Finally, for $r=4$, the sum is $379 + ({}^7 C_4 \cdot 2^4) = 379 + 560 = 939$. This confirms that the positive powers of $x$ exist specifically for $r = 0, 1, 2, 3,$ and $4$.

To ensure $r=4$ is the last term with a positive power of $x$, the value of $\frac{7n}{n+5}$ must satisfy the inequality $4 < \frac{7n}{n+5} < 5$. Solving the first part, $4 < \frac{7n}{n+5}$ implies $4n+20 < 7n$, which simplifies to $3n > 20$, or $n > 6.66$. Solving the second part, $\frac{7n}{n+5} < 5$ implies $7n < 5n+25$, which simplifies to $2n < 25$, or $n < 12.5$. The integer values of $n$ that satisfy these conditions are $7, 8, 9, 10, 11,$ and $12$. The sum of these possible values is $7 + 8 + 9 + 10 + 11 + 12 = 57$.

The term independent of $x$ in the product of $(1-x^2+3x^3)$ and $\left(\frac{5}{2}x^3-\frac{1}{5x^2}\right)^{11}$ is determined by first identifying the general term of the binomial expansion. Using the binomial theorem, the general term is given by $T_{r+1}={}^{11}C_r \cdot \left(\frac{5}{2}x^3\right)^{11-r} \cdot \left(-\frac{1}{5x^2}\right)^r$, which simplifies to:

$T_{r+1} = {}^{11}C_r \cdot \left(\frac{5}{2}\right)^{11-r} \cdot \left(-\frac{1}{5}\right)^r \cdot x^{33-5r}$

Because we are multiplying this expression by $(1-x^2+3x^3)$, we seek the constant term by checking which values of $r$ result in an $x^0$ term. This occurs when the power of the binomial expansion $x^{33-5r}$ combines with the powers $x^0$, $x^2$, and $x^3$ from the trinomial to sum to zero. Specifically, the binomial exponent must be $0$ (for the factor 1), $-2$ (for the factor $-x^2$), or $-3$ (for the factor $3x^3$). Testing these conditions leads to $33-5r=0$, $33-5r=-2$, and $33-5r=-3$, yielding $r=6.6$, $r=7$, and $r=7.2$ respectively.

Only $r=7$ provides an integer result, indicating that the constant term arises exclusively from the product of $-x^2$ and the binomial term where the power of $x$ is $-2$. Consequently, we calculate the independent term as $-1 \cdot {}^{11}C_7 \cdot \left(\frac{5}{2}\right)^{11-7} \cdot \left(-\frac{1}{5}\right)^7$. This simplifies to $- \frac{11 \cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2 \cdot 1} \cdot \frac{5^4}{2^4} \cdot \left(-\frac{1}{5^7}\right)$, which further reduces to $\frac{330}{16 \cdot 125}$. Final simplification results in $\frac{330}{2000}$, which is $\frac{33}{200}$.

The binomial expansion of $(a+b)^n$ relies on the general term formula $T_{k+1} = {}^{n}C_k a^{n-k} b^k$. For the specific expression $\left(2 x^{\frac{1}{5}} - x^{-\frac{1}{5}}\right)^{15}$, the general term is defined as $T_{k+1} = {}^{15}C_k \left(2 x^{\frac{1}{5}}\right)^{15-k} \left(-x^{-\frac{1}{5}}\right)^k$. This simplifies to ${}^{15}C_k 2^{15-k} (-1)^k x^{\frac{15-2k}{5}}$.

To identify the coefficient $m$ of $x^{-1}$, we set the exponent of $x$ equal to $-1$, resulting in $\frac{15-2k}{5} = -1$. This simplifies to $15-2k = -5$, yielding $2k = 20$ and $k=10$. Substituting $k=10$ into the expression, we find $m = {}^{15}C_{10} 2^{15-10} (-1)^{10} = {}^{15}C_{10} 2^5$. Similarly, to find the coefficient $n$ of $x^{-3}$, we set $\frac{15-2k}{5} = -3$, which leads to $15-2k = -15$, meaning $2k = 30$ and $k=15$. Consequently, $n = {}^{15}C_{15} 2^{15-15} (-1)^{15} = 1 \cdot 1 \cdot (-1) = -1$.

Given the product $m n^2$, we substitute the values to obtain $({}^{15}C_{10} 2^5) (-1)^2 = {}^{15}C_{10} 2^5$. By the symmetry property of combinations, ${}^{15}C_{10} = {}^{15}C_{5}$, so the expression becomes ${}^{15}C_5 2^5$. Matching this to the form ${}^{15}C_r 2^r$ demonstrates that $r$ is equal to 5.

Recognizing the expression as a finite geometric series with first term $(5+x)^{500}$ and common ratio $\frac{x}{5+x}$ provides a direct path to simplification. Because there are 501 terms total, the sum is determined by the formula $S = a \frac{1-r^n}{1-r}$. Substituting the terms yields $S = (5+x)^{500} \frac{1 - (\frac{x}{5+x})^{501}}{1 - \frac{x}{5+x}}$, which reduces to $S = \frac{1}{5} ((5+x)^{501} - x^{501})$ after algebraic cancellation.

To isolate the coefficient of $x^{101}$, we examine the binomial expansion of $(5+x)^{501}$, where the term containing $x^{101}$ is given by ${}^{501}C_{101} 5^{501-101} x^{101}$. Applying the external factor of $\frac{1}{5}$ to this binomial term results in $\frac{1}{5} \cdot {}^{501}C_{101} 5^{400}$, which simplifies to ${}^{501}C_{101} 5^{399}$.

Applying Pascal's Identity, ${}^nC_r + {}^nC_{r+1} = {}^{n+1}C_{r+1}$, provides the foundation for this calculation. Since ${}^{40}C_0$ equals 1, replacing it with ${}^{41}C_0$ initiates a telescoping pattern where ${}^{41}C_0 + {}^{41}C_1$ becomes ${}^{42}C_1$, which then combines with ${}^{42}C_2$ to form ${}^{43}C_2$, and this progression continues through the entire sequence. This cumulative process simplifies the expression to ${}^{61}C_{20}$. Equating this result to $\frac{m}{n} {}^{60}C_{20}$ leads to the following relationship:

$\frac{61!}{20! 41!} = \frac{m}{n} \cdot \frac{60!}{20! 40!}$

Simplifying this equation reveals that $\frac{61}{41} = \frac{m}{n}$, identifying $m=61$ and $n=41$ as the coprime pair. Adding these two values yields 102.

Calculating the remainder of a large power like $3^{2022}$ when divided by 5 is most efficiently handled by identifying the repeating cycle of its values modulo 5. The powers of 3 exhibit a clear periodic pattern: $3^1 \equiv 3$, $3^2 \equiv 4$, $3^3 \equiv 2$, and $3^4 \equiv 1$. This four-step sequence of remainders-3, 4, 2, and 1-repeats indefinitely as the exponent increases.

To determine the remainder for $3^{2022}$, we divide the exponent 2022 by the cycle length of 4. Since 2022 divided by 4 leaves a remainder of 2, the value $3^{2022}$ is equivalent to $3^2 \pm od 5$. Because $3^2 = 9$, and dividing 9 by 5 leaves a remainder of 4, the final remainder is 4.

Simplifying the expression $11^{1011} + 1011^{11}$ modulo 9 begins by reducing the bases, as $11 \equiv 2 \pm od 9$ and $1011 \equiv 3 \pm od 9$. The expression then becomes:
$2^{1011} + 3^{11} \equiv 2^{1011} + 0 \pm od 9$
Since $3^{11}$ contains $3^2$, it is divisible by 9, and $2^{1011}$ can be rewritten by observing that $2^3 = 8 \equiv -1 \pm od 9$, which leads to $(2^3)^{337} = 8^{337} \equiv (-1)^{337} \equiv -1 \pm od 9$. Adding 9 to the result $-1$ gives a final remainder of 8.

The term independent of $t$ in a binomial expansion is identified by determining the specific value of $r$ for which the net exponent of $t$ becomes zero. For the expansion $\left( t^{2} x^{\frac{1}{5}} + t^{-1} (1-x)^{\frac{1}{10}} \right)^{15}$, the general term is given by $T_{r+1} = {}^{15}C_{r} \left( t^{2} x^{\frac{1}{5}} \right)^{15-r} \left( t^{-1} (1-x)^{\frac{1}{10}} \right)^{r}$. Simplifying the expression by combining the powers of $t$ yields $t^{30-2r} \cdot t^{-r} = t^{30-3r}$, and setting the exponent $30-3r = 0$ reveals that $r=10$ corresponds to the term independent of $t$. Substituting this value back into the general term expression results in $T_{11} = {}^{15}C_{10} x^{1} (1-x)^{1}$.