Binomial Theorem – JEE Main MathematicsPractice Questions & PYQs

The sum of the coefficients in any binomial expansion of $(x+y)^{n}$ is equivalent to the expression evaluated at $x=1$ and $y=1$, which simplifies to $2^{n}$. Given that this sum is 4096, we solve the equation $2^{n} = 4096$ to find that $n$ must be 12, since $2^{12}$ equals 4096. When the power $n$ is an even number, the largest binomial coefficient always occurs at the middle term, which is calculated using the combination formula ${}^{n}C_{n/2}$. Substituting $n=12$ into this expression gives ${}^{12}C_{6}$, which results in 924.

Since the remainder of $x$ divided by $4$ is $3$, we can express $x$ as $4p+3$, which makes the base $2020+x = 2020+4p+3 = 4(505+p)+3$, simplifying to the form $4k+3$ for some integer $k$. Squaring this base gives $(4k+3)^2 = 16k^2+24k+9$, which can be written as $8(2k^2+3k+1)+1$, demonstrating that the square of the base is always congruent to $1$ modulo $8$. Because the exponent $2022$ is an even number, we can rewrite the expression as $((4k+3)^2)^{1011}$, which simplifies to $1^{1011}$ and results in a remainder of $1$.

The general term in the expansion of a binomial $(a+b)^n$ is defined by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$, where $0 \leq r \leq n$. By substituting $a = (2^2)^{1/4} = 2^{1/2}$, $b = 5^{1/6}$, and $n = 120$, the expression for the general term becomes $T_{r+1} = \binom{120}{r} 2^{\frac{120-r}{2}} 5^{\frac{r}{6}}$. For a term to be rational, the exponents of 2 and 5 must be integers. This requires both $\frac{120-r}{2}$ and $\frac{r}{6}$ to result in integer values.

Since $\frac{120-r}{2}$ is an integer for any even value of $r$, and $\frac{r}{6}$ is an integer whenever $r$ is a multiple of 6, both conditions are satisfied as long as $r$ is a multiple of 6. Within the range $0 \leq r \leq 120$, the permissible values for $r$ are $0, 6, 12, \dots, 120$. These values take the form $r = 6k$, where $k$ ranges from $0$ to $20$. Consequently, there are exactly 21 such terms.

To find the sum of coefficients with odd indices, start by substituting $x=1$ and $x=-1$ into the given polynomial $(1+x+2x^2)^{20}$. Setting $x=1$ yields:
$4^{20} = a_0 + a_1 + a_2 + \dots + a_{40}$
Setting $x=-1$ yields:
$2^{20} = a_0 - a_1 + a_2 - a_3 + \dots + a_{40}$
Subtracting the second expression from the first cancels out all terms with even indices and leaves twice the sum of odd-indexed coefficients:
$2(a_1 + a_3 + \dots + a_{39}) = 4^{20} - 2^{20}$
Dividing both sides by two, the sum up to $a_{39}$ is $2^{39} - 2^{19}$.

Since the question asks for the sum only up to $a_{37}$, identify the value of the $a_{39}$ term and remove it from the total sum. Applying the multinomial theorem, the coefficient $a_{39}$ is found by determining the combination of powers such that the exponent sum is $39$, which occurs at $20 \times 2^{19}$. Subtracting this value from the total sum produces the final expression:
$2^{39} - 2^{19} - 20(2^{19}) = 2^{19}(2^{20} - 21)$

This problem utilizes Vandermonde's Identity, which states that $\sum_{i=0}^{k} \binom{m}{i} \binom{n}{k-i} = \binom{m+n}{k}$. By applying the symmetry property of binomial coefficients, $\binom{n}{r} = \binom{n}{n-r}$, the terms within the summations can be reorganized to match this identity. The first expression $\sum_{i=0}^{9} \binom{9}{i} \binom{12}{12-k+i}$ is equivalent to $\sum_{i=0}^{9} \binom{9}{i} \binom{12}{k-i}$, which simplifies to $\binom{21}{k}$. Similarly, the second expression $\sum_{i=0}^{8} \binom{8}{i} \binom{13}{13-k+i}$ can be written as $\sum_{i=0}^{8} \binom{8}{i} \binom{13}{k-i}$, which also simplifies to $\binom{21}{k}$.

Summing these components yields the expression $A_k = 2 \binom{21}{k}$. To find $p$, we evaluate the difference $A_4 - A_3 = 2 \binom{21}{4} - 2 \binom{21}{3}$. Using the standard formula for combinations, $\binom{21}{4} = 5985$ and $\binom{21}{3} = 1330$. The expression becomes $2(5985 - 1330)$, which simplifies to $2(4655)$, resulting in $A_4 - A_3 = 9310$. Given the equation $190p = 9310$, dividing both sides by 190 confirms that $p = 49$.

Begin by factoring $a$ out of each denominator to rewrite the sum as $\frac{1}{a} \sum_{r=1}^{n} (1 - \frac{rb}{a})^{-1}$. Using the binomial expansion $(1-x)^{-1} \approx 1+x+x^2$ for small values of $x$, we substitute $\frac{rb}{a}$ for $x$ to approximate each term as $1 + \frac{rb}{a} + \frac{r^2b^2}{a^2}$. Summing these expressions leads to $\frac{1}{a} [n + \frac{b}{a}\frac{n(n+1)}{2} + \frac{b^2}{a^2}\frac{n(n+1)(2n+1)}{6}]$. Since we need the coefficient $\gamma$ of the $n^3$ term, we focus on the highest-order contribution from $\sum_{r=1}^{n} r^2$, which is $\frac{n^3}{3}$, yielding $\gamma = \frac{b^2}{3a^3}$.

Recognizing the algebraic identity for the difference of cubes, $(1-x)(1+x+x^2) = 1-x^3$, allows us to rewrite the expression $(1-x)^{101}(x^2+x+1)^{100}$ as $(1-x)(1-x)^{100}(x^2+x+1)^{100}$, which simplifies to $(1-x)(1-x^3)^{100}$. Expanding this product gives $(1-x^3)^{100} - x(1-x^3)^{100}$. We seek the coefficient of $x^{256}$ in this expression. Since $(1-x^3)^{100}$ contains only powers of $x$ that are multiples of $3$, it cannot contribute to the $x^{256}$ term. We therefore focus on the second part, $-x(1-x^3)^{100}$, and identify the coefficient of $x^{255}$ within the expansion of $(1-x^3)^{100}$. By the binomial theorem, the general term is ${}^{100}C_{r}(-x^3)^r$, and setting $3r = 255$ gives $r = 85$. The specific term for $r=85$ is ${}^{100}C_{85}(-1)^{85}x^{255} = -{}^{100}C_{85}x^{255}$. Multiplying this by the leading $-x$ results in $-x(-{}^{100}C_{85}x^{255}) = {}^{100}C_{85}x^{256}$. Applying the symmetry property ${}^{n}C_{r} = {}^{n}C_{n-r}$, this coefficient is equivalent to ${}^{100}C_{15}$.

For $3^{n} + 7^{n}$ to be a multiple of 10, the expression must be congruent to 0 modulo 10. Noting that $7$ is equivalent to $-3$ modulo 10 allows the expression to be rewritten as $3^{n} + (-3)^{n}$. If $n$ is an even integer, $(-3)^{n}$ becomes positive, resulting in $2 \cdot 3^{n}$, which produces last digits of 2, 4, 6, or 8 and never 0. However, when $n$ is an odd integer, $(-3)^{n}$ remains negative, resulting in $3^{n} - 3^{n}$, which equals 0. Thus, the condition is satisfied if and only if $n$ is odd.

The set of two-digit numbers ranges from 10 to 99, providing a total of 90 integers. Because this range contains an equal number of odd and even values, exactly half of these numbers are odd. Counting every odd integer in this sequence, starting from 11 and ending at 99, results in 45 possible values for $n$.

The expression $\sum_{r=0}^{6} ({}^{6}C_{r} \cdot {}^{6}C_{6-r})$ is a classic application of Vandermonde's Identity, which relates the sum of products of binomial coefficients to a single binomial coefficient. Specifically, this sum represents the coefficient of $x^6$ in the expansion of $(1+x)^6 \cdot (1+x)^6$, which is equivalent to the expansion of $(1+x)^{12}$.

Applying this property to identify the coefficient of $x^6$ in $(1+x)^{12}$ leads directly to the combination ${}^{12}C_6$. Calculating this value requires evaluating the expression $\frac{12!}{6!(12-6)!}$, which simplifies to $\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$. Completing the division results in a final value of 924.

Tackling a sum involving $r^{2} \, {}^{n}C_{r}$ is most straightforward when you rewrite $r^{2}$ as the expression $r(r-1) + r$. This algebraic manipulation allows you to utilize the property $r \, {}^{n}C_{r} = n \, {}^{n-1}C_{r-1}$, which when applied twice yields $r(r-1) \, {}^{n}C_{r} = n(n-1) \, {}^{n-2}C_{r-2}$. By breaking the original summation into two separate parts, you transform the problem from a complex series into the evaluation of binomial identities that are essentially powers of two.

When applying this to the expansion where $n=20$, the sum $\sum_{r=0}^{20} r^{2} \, {}^{20}C_{r}$ splits into $\sum_{r=2}^{20} 20 \times 19 \, {}^{18}C_{r-2} + \sum_{r=1}^{20} 20 \, {}^{19}C_{r-1}$. Evaluating these sums using the identity $\sum_{k=0}^{m} {}^{m}C_{k} = 2^{m}$, the first component results in $380 \times 2^{18}$, while the second component results in $20 \times 2^{19}$. Converting $20 \times 2^{19}$ into $40 \times 2^{18}$ allows you to combine the terms, yielding $380 \times 2^{18} + 40 \times 2^{18}$, which simplifies to $420 \times 2^{18}$.

Vandermonde's Identity allows us to simplify these summations by recognizing them as coefficients in binomial expansions. The first summation $\sum_{i=0}^{k} \binom{10}{i} \binom{15}{k-i}$ corresponds to the coefficient of $x^k$ in the product $(1+x)^{10}(1+x)^{15}$, which simplifies to $\binom{25}{k}$. Similarly, the second summation $\sum_{i=0}^{k+1} \binom{12}{i} \binom{13}{k+1-i}$ corresponds to the coefficient of $x^{k+1}$ in the product $(1+x)^{12}(1+x)^{13}$, which simplifies to $\binom{25}{k+1}$. Combining these two results, we get $\binom{25}{k} + \binom{25}{k+1}$, which by Pascal's Identity equals $\binom{26}{k+1}$. Given that the expression $\binom{n}{r}$ is non-zero only when $n \ge r \ge 0$, the coefficient $\binom{26}{k+1}$ is non-zero for $k+1 \le 26$, implying that the maximum integer value for $k$ is 25.

Simplifying the expression within the parentheses is the most efficient path, as the first part $\frac{x+1}{x^{2/3} - x^{1/3} + 1}$ equates to $(x^{1/3} + 1)$ because the numerator represents a sum of cubes factorization. Similarly, the second part $\frac{x-1}{x - x^{1/2}}$ simplifies to $1 + x^{-1/2}$ after dividing both numerator and denominator by $\sqrt{x}$, reducing the entire base to $(x^{1/3} - x^{-1/2})^{10}$. The general term for this binomial expansion is defined as $T_{r+1} = {}^{10}C_{r} (x^{1/3})^{10-r} (-1)^r (x^{-1/2})^r$. To isolate the term independent of $x$, the total exponent of $x$ must be zero, requiring the condition $\frac{10-r}{3} - \frac{r}{2} = 0$. Solving this equation gives $20 - 2r = 3r$, which results in $r = 4$. Substituting this index into the binomial coefficient yields $T_5 = {}^{10}C_4$, which calculates to 210.

The expansion of $(x+\sqrt{x^2-1})^6+(x-\sqrt{x^2-1})^6$ follows the binomial identity for $(x+a)^n+(x-a)^n$, where the terms containing odd powers of the radical part cancel each other out, leaving only the even-indexed terms. This simplification yields the expansion $2[{}^6C_0 x^6 + {}^6C_2 x^4(x^2 - 1) + {}^6C_4 x^2(x^2 - 1)^2 + {}^6C_6(x^2 - 1)^3]$.

Evaluating these components, the term $2 \cdot 15 \cdot x^4(x^2 - 1)$ simplifies to $30x^6 - 30x^4$, while $2 \cdot 15 \cdot x^2(x^4 - 2x^2 + 1)$ results in $30x^6 - 60x^4 + 30x^2$, and $2 \cdot 1 \cdot (x^6 - 3x^4 + 3x^2 - 1)$ produces $2x^6 - 6x^4 + 6x^2 - 2$. Aggregating the coefficients reveals the value of $\alpha$ as $-30 - 60 - 6 = -96$ and the value of $\beta$ as $30 + 6 = 36$. Consequently, the difference $\alpha - \beta$ evaluates to $-96 - 36$, resulting in $-132$.

In any binomial expansion of the form $(\alpha x^a + \beta x^b)^n$, the term independent of $x$ occurs where the net exponent of the variable is zero. For the expansion $(\alpha x^{\frac{1}{9}}+\beta x^{-\frac{1}{6}})^{10}$, the general term is given by $\binom{10}{r} \alpha^{10-r} \beta^r x^{\frac{10-r}{9} - \frac{r}{6}}$. Setting the exponent equal to zero yields the equation $\frac{10-r}{9} = \frac{r}{6}$, which simplifies to $20-2r = 3r$, solving to $r=4$. This confirms that the fifth term, involving $\binom{10}{4} \alpha^6 \beta^4$, is the one devoid of the variable $x$.

The value of this independent term is $\binom{10}{4} \alpha^6 \beta^4$. To maximize this quantity subject to the constraint $\alpha^3 + \beta^2 = 4$, we apply the Arithmetic Mean-Geometric Mean inequality by splitting the terms to align with the exponents in the product $\alpha^6 \beta^4$. We evaluate:

$\frac{ \frac{\alpha^3}{2} + \frac{\alpha^3}{2} + \frac{\beta^2}{2} + \frac{\beta^2}{2} }{4} \geq \left( \frac{\alpha^6 \beta^4}{16} \right)^{\frac{1}{4}}$

Substituting the known sum of 4 into the left side yields $1 \geq \left( \frac{\alpha^6 \beta^4}{16} \right)^{\frac{1}{4}}$, which implies $\alpha^6 \beta^4 \leq 16$. Using the binomial coefficient $\binom{10}{4} = 210$, the maximum value of the independent term is $210 \times 16 = 3360$. Setting this result equal to the given expression $10k$, we find $k = 336$.

The expression is a summation of terms in the form $x^r(1+x)^{10-r}$, where the coefficient of $x^7$ is determined by extracting $x^{7-r}$ from each expansion $(1+x)^{10-r}$, resulting in the sequence $\binom{10}{7} + \binom{9}{6} + \binom{8}{5} + \binom{7}{4} + \binom{6}{3} + \binom{5}{2} + \binom{4}{1} + \binom{3}{0}$. By applying the symmetry property $\binom{n}{k} = \binom{n}{n-k}$, this sum simplifies to $\binom{10}{3} + \binom{9}{3} + \binom{8}{3} + \binom{7}{3} + \binom{6}{3} + \binom{5}{3} + \binom{4}{3} + \binom{3}{3}$. Using the hockey-stick identity $\sum_{i=k}^{n} \binom{i}{k} = \binom{n+1}{k+1}$, the entire series reduces to $\binom{11}{4}$, which yields 330.

The general term for the binomial expansion of $(x^m + x^{-2})^{22}$ is expressed as $T_{r+1} = \binom{22}{r} (x^m)^{22-r} (x^{-2})^r$, which simplifies to $T_{r+1} = \binom{22}{r} x^{22m - mr - 2r}$. Because the coefficient of $x$ is 1540, we search for values of $r$ where $\binom{22}{r} = 1540$. Using the symmetry property of binomial coefficients, we find that this condition is satisfied when $r=3$ or $r=19$.

To determine the value of $m$, we set the exponent of $x$ in the general term equal to 1, leading to the equation $22m - mr - 2r = 1$. Rearranging this to solve for $m$ yields $m = \frac{2r+1}{22-r}$. Testing the first possibility, $r=3$, gives $m = \frac{2(3)+1}{22-3} = \frac{7}{19}$, which is not a natural number. Substituting the second possibility, $r=19$, results in $m = \frac{2(19)+1}{22-19} = \frac{39}{3}$, yielding the natural number 13.

The general term in the binomial expansion of $(3^{\frac{1}{2}} + 5^{\frac{1}{8}})^n$ is given by $T_{r+1} = \binom{n}{r} (3)^{\frac{n-r}{2}} (5)^{\frac{r}{8}}$, where $r$ ranges from $0$ to $n$. For each term to be an integer, the exponent of 5 must be a whole number, necessitating that $r$ is a multiple of 8, and the exponent of 3 must be a whole number, requiring $n-r$ to be even. Since any multiple of 8 is already even, selecting $n$ as a multiple of 8 ensures that $n-r$ remains even whenever $r$ is a multiple of 8. With exactly 33 integral terms, $r$ takes values in the arithmetic progression $\{0, 8, 16, \dots, 8k\}$, where the 33rd value corresponds to $k=32$. This leads to $r = 32 \times 8 = 256$, which is the minimum value required for $n$ to accommodate these 33 terms.

Calculating the fractional part of $\frac{3^{200}}{8}$ involves determining the remainder left when $3^{200}$ is divided by 8, then dividing that remainder by 8. By rewriting $3^{200}$ as $(3^2)^{100}$, we obtain $9^{100}$, which can be expressed as $(8+1)^{100}$. Because every term in the binomial expansion of $(8+1)^{100}$ contains a factor of 8 except for the final term of $1^{100}$, the numerator leaves a remainder of 1 when divided by 8. Therefore, the fractional part is $\frac{1}{8}$.

The binomial coefficients in an expansion are determined by the sequence ${}^{n}C_{k}$, and when three consecutive terms are provided in a specific ratio, we can establish relationships using the property $\frac{{}^{n}C_{k}}{{}^{n}C_{k-1}} = \frac{n-k+1}{k}$. Given the ratio ${}^{n}C_{r-1}:{}^{n}C_{r}:{}^{n}C_{r+1} = 2:5:12$, we can isolate two separate equations to represent the relationships between $n$ and $r$.

From the first segment of the ratio, $\frac{{}^{n}C_{r-1}}{{}^{n}C_{r}} = \frac{2}{5}$, we apply the ratio formula to get $\frac{r}{n-r+1} = \frac{2}{5}$, which rearranges to $7r = 2n + 2$. Similarly, from the second segment, $\frac{{}^{n}C_{r}}{{}^{n}C_{r+1}} = \frac{5}{12}$, we use the formula to obtain $\frac{r+1}{n-r} = \frac{5}{12}$, simplifying to $17r = 5n - 12$. Solving this system of equations by eliminating $r$ leads directly to the solution $n = 118$.

The relationship between coefficients of a polynomial expansion can be revealed by substituting the variable with its reciprocal scaled by a constant factor. Given $(2x^2 + 3x + 4)^{10} = \sum_{r=0}^{20} a_r x^r$, substituting $x$ with $\frac{2}{x}$ yields the expression $(\frac{8}{x^2} + \frac{6}{x} + 4)^{10} = \sum_{r=0}^{20} a_r (\frac{2}{x})^r$. Simplifying the left side results in $(\frac{4x^2 + 6x + 8}{x^2})^{10}$, which is equal to $\frac{2^{10}(2x^2 + 3x + 4)^{10}}{x^{20}}$. By substituting the original summation back into this expression, we obtain:

$\frac{2^{10}}{x^{20}} \sum_{k=0}^{20} a_k x^k = \sum_{r=0}^{20} a_r 2^r x^{-r}$

Expanding both sides reveals the identity $\sum_{k=0}^{20} a_k 2^{10} x^{k-20} = \sum_{r=0}^{20} a_r 2^r x^{-r}$. To find the ratio $\frac{a_7}{a_{13}}$, we compare the coefficients of the matching power $x^{-7}$. On the right side, this corresponds to $r=7$, and on the left side, this corresponds to $k-20 = -7$, which implies $k=13$. Equating the coefficients $a_7 \cdot 2^7$ and $a_{13} \cdot 2^{10}$ leads to:

$\frac{a_7}{a_{13}} = \frac{2^{10}}{2^7} = 2^3 = 8$