Binomial Theorem – JEE Main MathematicsPractice Questions & PYQs

Each term in the series $\frac{1}{1!25!} + \frac{1}{3!23!} + \frac{1}{5!21!} + \dots$ resembles a binomial coefficient structure if we introduce the factor $26!$. By multiplying the entire sum by $\frac{26!}{26!}$, the expression transforms into $\frac{1}{26!} \sum \binom{26}{r}$, where $r$ takes on the odd values 1, 3, 5, up to 25 for the 13 given terms.

The sum of binomial coefficients with odd indices, specifically $\binom{n}{1} + \binom{n}{3} + \dots$, follows the identity $2^{n-1}$. Applying this to our case where $n=26$, the sum of these coefficients equals $2^{25}$. Therefore, the entire series $S$ simplifies to the expression $\frac{2^{25}}{26!}$.

Substituting this result into the given equation $13 S = \frac{2^k}{n!}$ yields $\frac{13 \cdot 2^{25}}{26!} = \frac{2^k}{n!}$. Since $26! = 26 \times 25!$, the left side becomes $\frac{13 \cdot 2^{25}}{26 \cdot 25!}$, which reduces to $\frac{2^{24}}{25!}$. Comparing the two sides of the equation, we identify $k = 24$ and $n = 25$, and adding these values together results in 49.

The identity $1-x^3 = (1-x)^3 + 3x(1-x)$ serves as the foundation for expanding the expression $(1-x^3)^{10}$ using the binomial theorem. Substituting this identity into the expression results in $[(1-x)^3 + 3x(1-x)]^{10}$, which expands as:

$\sum_{r=0}^{10} \binom{10}{r} ((1-x)^3)^{10-r} (3x(1-x))^r$

This summation simplifies to $\sum_{r=0}^{10} \binom{10}{r} 3^r x^r (1-x)^{30-2r}$, directly identifying the coefficients as $a_r = \binom{10}{r} 3^r$. Using this general form, we can calculate $a_9 = \binom{10}{9} 3^9 = 10 \cdot 3^9$ and $a_{10} = \binom{10}{10} 3^{10} = 3^{10}$. The ratio $\frac{a_9}{a_{10}}$ becomes $\frac{10 \cdot 3^9}{3^{10}}$, which simplifies to $\frac{10}{3}$. Multiplying this value by $9$ yields $\frac{9 a_9}{a_{10}} = 9 \cdot \frac{10}{3} = 30$.

The behavior of any term in a binomial expansion is determined by the general term formula, which allows for isolating the coefficient of a specific power of $x$ by equating the variable's net exponent to the desired value. In the expansion of $\left(\frac{1}{x^{3}}-x^{4}\right)^{n}$, the general term is defined as $T_{r+1} = {}^{n}C_{r} \left(\frac{1}{x^{3}}\right)^{n-r} (-x^{4})^{r}$, which simplifies upon consolidation to ${}^{n}C_{r} (-1)^{r} x^{7r-3n}$. Identifying the coefficients for $x^{7}$ and $x^{14}$ involves solving the equations $7r-3n = 7$ and $7r-3n = 14$ for $r$, yielding $r_1 = \frac{3n+7}{7}$ and $r_2 = \frac{3n+14}{7}$, where the latter is simply $r_1 + 1$.

Because the sum of these coefficients is given as zero, the expression ${}^{n}C_{r_1} (-1)^{r_1} + {}^{n}C_{r_1+1} (-1)^{r_1+1} = 0$ must hold true. Factoring out $(-1)^{r_1}$ demonstrates that ${}^{n}C_{r_1} - {}^{n}C_{r_1+1} = 0$, meaning the two binomial coefficients must be equal. Using the symmetry property of binomial coefficients where ${}^{n}C_k = {}^{n}C_{k+1}$ occurs only when $k + (k + 1) = n$, we arrive at the equation $2r_1 + 1 = n$. Substituting the expression $r_1 = \frac{3n+7}{7}$ into this equality provides $2\left(\frac{3n+7}{7}\right) + 1 = n$. Multiplying throughout by $7$ yields $6n + 14 + 7 = 7n$, which simplifies to $n = 21$.

The divisibility of an expression involving exponents with identical odd indices can be determined by grouping terms according to the algebraic identity $x^n + y^n = (x+y)(x^{n-1} - x^{n-2}y + \dots + y^{n-1})$. Applying this to the sum $25^{13} + 20^{13} + 8^{13} + 3^{13}$, we can rearrange the terms into two pairs: $(25^{13} + 3^{13})$ and $(20^{13} + 8^{13})$. Since $13$ is odd, each binomial pair is divisible by the sum of its respective bases; specifically, $25+3=28$ and $20+8=28$. Because both sums are multiples of $28$, which is itself divisible by $7$, the entire expression is a multiple of $7$, confirming that the first statement is true.

Evaluating the integral part of $(7+4\sqrt{3})^{25}$ involves using the conjugate $(7-4\sqrt{3})^{25}$ and binomial expansion. Let $I$ be the integral part and $f$ be the fractional part, such that $(7+4\sqrt{3})^{25} = I + f$, with $0 < f < 1$. Given that $7-4\sqrt{3} \approx 0.07$, raising this value to the $25$th power results in a positive fraction $f'$ such that $0 < f' < 1$. When adding $(7+4\sqrt{3})^{25}$ and $(7-4\sqrt{3})^{25}$, the binomial expansion ensures that all terms involving odd powers of $\sqrt{3}$ cancel out, leaving an even integer. Therefore, $(I+f) + f'$ must equal an even integer, which forces $f+f'$ to equal $1$ because $0 < f+f' < 2$. Since $I + f + f'$ is even and $f+f' = 1$, it follows that $I + 1$ is even, proving that $I$ must be an odd number and the second statement is true.

Expanding $(1-2x)^{26}$ via the binomial theorem provides the general term $\binom{26}{r}(-2)^r x^r$. Distributing the trinomial $(ax^2+bx+c)$ across this expansion allows us to construct a system of linear equations based on the specified coefficients. For the coefficient of $x$, we collect terms from $b \cdot \binom{26}{0}(-2)^0 x^1$ and $c \cdot \binom{26}{1}(-2)^1 x^1$, which yields:
$b - 52c = -56$
For the coefficient of $x^2$, we sum the contributions from $a$, $b$, and $c$ terms, resulting in:
$a - 52b + 1300c = 0$
The coefficient of $x^3$ involves contributions from all three parts of the trinomial, which simplifies to:
$-52a + 1300b - 20800c = 0$

Dividing the third equation by 52 simplifies it to $-a + 25b - 400c = 0$. We can now solve the system by expressing $b$ as $52c - 56$ and substituting this into the second and third equations. Substituting into the second equation gives $a = 1404c - 2912$, while substituting into the simplified third equation gives $a = 900c - 1400$. Equating these two expressions leads to $1404c - 2912 = 900c - 1400$, which simplifies to $504c = 1512$, resulting in $c = 3$. Using this value, we find $b = 52(3) - 56 = 100$ and $a = 900(3) - 1400 = 1300$. The final sum is $a + b + c = 1300 + 100 + 3 = 1403$.

The sum $S = (1+x) + 2(1+x)^2 + \dots + 100(1+x)^{100}$ is an arithmetico-geometric series, which can be evaluated by subtracting $(1+x)S$ from $S$ to create a telescoping summation. This subtraction simplifies the expression to $-xS = (1+x) + (1+x)^2 + \dots + (1+x)^{100} - 100(1+x)^{101}$, where the geometric component evaluates to $\frac{(1+x)^{101}-(1+x)}{x}$. Rearranging terms reveals $S = \frac{100(1+x)^{101}}{x} - \frac{(1+x)^{101}}{x^2} + \frac{1+x}{x^2}$.

The coefficient of $x^{48}$ is found by examining the binomial expansion of $(1+x)^{101} = \sum \binom{101}{k}x^k$. The expression $\frac{100(1+x)^{101}}{x}$ provides $100$ times the coefficient of $x^{49}$, which is $100 \binom{101}{49}$, whereas $-\frac{(1+x)^{101}}{x^2}$ contributes the negative of the coefficient of $x^{50}$, resulting in $-\binom{101}{50}$. Combining these components yields the final value $100 \binom{101}{49} - \binom{101}{50}$.

The general term in the binomial expansion of $(2x^2 + \frac{1}{x})^{10}$ is given by
$T_{r+1} = {}^{10}C_{r} (2x^2)^{10-r} \left(\frac{1}{x}\right)^r$
By separating the numerical coefficients and powers of $x$, this expression becomes $T_{r+1} = {}^{10}C_{r} (2)^{10-r} x^{20-2r} x^{-r}$, which simplifies further to ${}^{10}C_{r} (2)^{10-r} x^{20-3r}$. Since the goal is to find the coefficient of $x^2$, the exponent of $x$ must satisfy $20 - 3r = 2$, which leads to $3r = 18$ or $r = 6$. Substituting this value back into the expression, the coefficient is calculated as ${}^{10}C_{6} (2)^{10-6}$, which is ${}^{10}C_{6} (2)^4$. Evaluating these terms, ${}^{10}C_{6}$ equals $210$ and $2^4$ equals $16$, resulting in a final value of $3360$.

Simplifying the sum of two adjacent reciprocals of binomial coefficients, $\frac{1}{{}^{n}C_r} + \frac{1}{{}^{n}C_{r+1}}$, reveals an elegant identity: $\frac{n+1}{n \cdot {}^{n-1}C_r}$. Applying this to the provided product where $n=15$ transforms each factor into $\frac{16}{15 \cdot {}^{14}C_r}$. Because there are 13 such factors in the product, ranging from $r=0$ to $r=12$, the entire expression becomes $\left( \frac{16}{15} \right)^{13} \cdot \frac{1}{{}^{14}C_0 {}^{14}C_1 \cdot s {}^{14}C_{12}}$. By comparing this result to the original equation, it is clear that $\alpha = \frac{16}{15}$. Consequently, calculating $30 \alpha$ involves multiplying 30 by $\frac{16}{15}$, which yields 32.

The binomial theorem states that the general term $T_{r+1}$ in the expansion of $(a+b)^n$ is defined by ${}^{n}C_{r} a^{n-r} b^{r}$. For the expression $\left(9x - \frac{1}{3\sqrt{x}}\right)^{18}$, the general term is expressed as ${}^{18}C_{r} (9x)^{18-r} \left(-\frac{1}{3}x^{-1/2}\right)^r$. Consolidating the variables reveals the power of $x$ as $x^{18-r} \cdot x^{-r/2}$, which simplifies to $x^{18 - \frac{3r}{2}}$.

To identify the term independent of $x$, the exponent of the variable must equal zero, requiring us to solve $18 - \frac{3r}{2} = 0$, which gives $r = 12$. Substituting this back into the formula yields a coefficient of ${}^{18}C_{12} \cdot 9^{18-12} \cdot \left(-\frac{1}{3}\right)^{12}$. Because $9^6$ is $(3^2)^6 = 3^{12}$ and $\left(-\frac{1}{3}\right)^{12} = \frac{1}{3^{12}}$, these powers of three cancel perfectly, leaving just the value of the combination ${}^{18}C_{12}$.

Calculating the combination ${}^{18}C_{12}$ is equivalent to ${}^{18}C_{6}$, which results in $\frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 18564$. The problem states that this coefficient is equal to $221k$. Dividing $18564$ by $221$ leads to $k = 84$.

The expansion of $(1+x^2)^2(1+x)^n$ is derived by expressing the first part as $(1+2x^2+x^4)$ and multiplying it by the binomial series for $(1+x)^n$. From this product, the coefficient of $x$ is $\binom{n}{1} = n$, the coefficient of $x^2$ is $\binom{n}{2} + 2$, and the coefficient of $x^3$ is $\binom{n}{3} + 2n$.

Arithmetic progression dictates that the middle term multiplied by $2$ equals the sum of the surrounding terms, expressed as $2\left(\binom{n}{2} + 2\right) = n + \left(\binom{n}{3} + 2n\right)$. Substituting the binomial formulas $\binom{n}{2} = \frac{n(n-1)}{2}$ and $\binom{n}{3} = \frac{n(n-1)(n-2)}{6}$ results in $n(n-1) + 4 = 3n + \frac{n(n-1)(n-2)}{6}$. Multiplying the entire equation by $6$ yields $6n(n-1) + 24 = 18n + n(n-1)(n-2)$, which simplifies to the cubic polynomial $n^3 - 9n^2 + 26n - 24 = 0$.

Factoring this equation produces $(n-2)(n-3)(n-4) = 0$, indicating that $n$ must be $2, 3,$ or $4$. These are the only natural numbers that satisfy the given condition, and their sum is $9$.

The binomial expansion of $(1+y)^n$ where $n$ is an even integer contains a single middle term at the position $\frac{n}{2} + 1$. For the expression $(1+\alpha x)^{26}$, the middle term is the 14th term, represented as ${}^{26}C_{13}(\alpha x)^{13}$, which results in a coefficient of ${}^{26}C_{13} \alpha^{13}$. Similarly, for the expression $(1-\alpha x)^{28}$, the middle term is the 15th term, represented as ${}^{28}C_{14}(-\alpha x)^{14}$, which results in a coefficient of ${}^{28}C_{14} \alpha^{14}$.

Equating these two coefficients and dividing both sides by $\alpha^{13}$-since $\alpha \neq 0$-isolates $\alpha$ as $\alpha = \frac{{}^{26}C_{13}}{{}^{28}C_{14}}$. Expanding these combinations using factorials gives $\frac{26!}{13!13!} \times \frac{14!14!}{28!}$, which simplifies to $\frac{14 \times 14}{28 \times 27}$ upon canceling out the common terms. Calculating this fraction results in $\frac{1}{2} \times \frac{14}{27}$, yielding the final value $\frac{7}{27}$.

The binomial theorem provides a straightforward approach to identifying the coefficient of $x^3$ in this series, specifically by leveraging the Hockey-stick identity, which posits that $\sum_{i=r}^{n} {}^{i}C_{r} = {}^{n+1}C_{r+1}$. By examining the expression $(1+x)^{3}+(1+x)^{4}+\dots+(1+x)^{99}$, the coefficient of $x^3$ from these terms is ${}^{3}C_{3}+{}^{4}C_{3}+\dots+{}^{99}C_{3}$. Adding the coefficient of $x^3$ from the final term $(1+kx)^{100}$, which is ${}^{100}C_{3}k^3$, the total coefficient of $x^3$ becomes the sum ${}^{3}C_{3}+{}^{4}C_{3}+\dots+{}^{99}C_{3}+{}^{100}C_{3}k^3$.

Since ${}^{3}C_{3}$ is equal to ${}^{4}C_{4}$, the series effectively transforms into ${}^{4}C_{4}+{}^{4}C_{3}+\dots+{}^{99}C_{3}$, which telescopes to ${}^{100}C_{4}$ according to the identity. This simplifies the total coefficient to ${}^{100}C_{4} + {}^{100}C_{3}k^3$. Equating this to the provided expression $\left(43 n + \frac{101}{4}\right) {}^{100}C_{3}$ and dividing the entire equation by ${}^{100}C_{3}$ yields the relationship $\frac{{}^{100}C_{4}}{{}^{100}C_{3}} + k^3 = 43 n + \frac{101}{4}$. The ratio $\frac{{}^{100}C_{4}}{{}^{100}C_{3}}$ simplifies to $\frac{97}{4}$. Substituting this back gives $\frac{97}{4} + k^3 = 43 n + \frac{101}{4}$, which simplifies further to $k^3 = 43 n + \frac{4}{4}$, or simply $k^3 = 43 n + 1$. Testing integer values for $n$ reveals that $n=5$ results in $k^3 = 43(5) + 1 = 216$, meaning $k=6$. With $p=6$ and $n=5$, the value of $p+n$ is 11.

The key to evaluating a series like $\frac{{}^{100}C_{50}}{51} + \frac{{}^{100}C_{51}}{52} + \dots + \frac{{}^{100}C_{100}}{101}$ is the identity $\frac{{}^{n}C_r}{r+1} = \frac{{}^{n+1}C_{r+1}}{n+1}$. Applying this property transforms each term in the sum into the form $\frac{{}^{101}C_{r+1}}{101}$, where $r$ ranges from $50$ to $100$.

Factoring out the constant $\frac{1}{101}$ leaves the summation of binomial coefficients from ${}^{101}C_{51}$ up to ${}^{101}C_{101}$. Because the sum of all binomial coefficients for a power of $101$ is $2^{101}$ and the series is perfectly symmetric, the sum of this latter half of the coefficients is exactly half of the total sum, which equals $2^{100}$. Multiplying this sum by the factor $\frac{1}{101}$ yields the final result of $\frac{2^{100}}{101}$.

The coefficients 1, 3, 3, 1 present in the expression correspond exactly to the binomial coefficients of the third row of Pascal's triangle, allowing the sum to be represented as ${}^{3}C_{3} \cdot {}^{30}C_{30-r} + {}^{3}C_{2} \cdot {}^{30}C_{31-r} + {}^{3}C_{1} \cdot {}^{30}C_{32-r} + {}^{3}C_{0} \cdot {}^{30}C_{33-r}$. This arrangement follows Vandermonde's Identity, which states that $\sum_{k=0}^{n} {}^{a}C_{k} \cdot {}^{b}C_{x-k} = {}^{a+b}C_{x}$, meaning the upper indices sum to $3+30=33$ and the lower indices sum to $33-r$, yielding ${}^{33}C_{33-r}$. Utilizing the symmetry property of binomial coefficients, ${}^{n}C_{k} = {}^{n}C_{n-k}$, we transform ${}^{33}C_{33-r}$ into ${}^{33}C_{33-(33-r)}$, which simplifies to ${}^{33}C_{r}$, demonstrating that $m$ must be 33.

This series functions as a geometric progression consisting of 1001 terms, with a first term of $(1+x)^{1000}$ and a common ratio of $\frac{x}{1+x}$. Utilizing the sum formula $S = a \frac{1 - r^n}{1 - r}$ allows the entire expression to simplify to $(1+x)^{1001} - x^{1001}$. Since we seek the coefficients for $x^{499}$ and $x^{500}$, the $x^{1001}$ term becomes negligible, leaving us to find the sum of $\binom{1001}{499}$ and $\binom{1001}{500}$ from the binomial expansion of $(1+x)^{1001}$. Invoking Pascal's identity, $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$, the result is $\binom{1002}{500}$.

The evaluation of sums involving binomial coefficients depends on the identity $\frac{\binom{n}{r}}{r+1} = \frac{1}{n+1} \binom{n+1}{r+1}$. This relationship enables the transformation of a series containing a denominator of $r+1$ into a structure that conforms to the standard binomial expansion theorem.

Substituting this identity into the definition of $P_n$ yields $P_n = \frac{1}{n+1} \sum_{r=0}^n \binom{n+1}{r+1}(-2)^r$. By performing a change of variables where $k = r+1$, the expression becomes $-\frac{1}{2(n+1)} \sum_{k=1}^{n+1} \binom{n+1}{k}(-2)^k$. Given that $(1-2)^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k}(-2)^k$, the summation simplifies to $(1-2)^{n+1} - \binom{n+1}{0}(-2)^0$, which results in $(-1)^{n+1} - 1$. Thus, the expression for $P_n$ is $\frac{1 - (-1)^{n+1}}{2(n+1)}$.

For an even index $2n$, the expression simplifies to $\frac{1 - (-1)^{2n+1}}{2(2n+1)}$, which equals $\frac{1 - (-1)}{2(2n+1)} = \frac{1}{2n+1}$. The reciprocal is therefore $\frac{1}{P_{2n}} = 2n+1$. The final summation $\sum_{n=1}^{25} (2n+1)$ constitutes an arithmetic series with 25 terms where the first term is 3 and the last term is 51, resulting in a total sum of $25 \times \frac{3+51}{2} = 675$.

To evaluate the sum, we utilize the combinatorial identity $\frac{1}{k+1} {}^{n}C_{k} = \frac{1}{n+1} {}^{n+1}C_{k+1}$. By applying this to the terms of the series where $n=12$, we can rewrite each term $\frac{2^{k+1}}{k+1} {}^{12}C_{k}$ as $\frac{2^{k+1}}{13} {}^{13}C_{k+1}$. Substituting this into the original expression, the factor of $26$ outside the parentheses simplifies with the denominator $13$ to leave a coefficient of $2$, transforming the summation into $2 \sum_{r \in \{3, 5, \dots, 13\}} 2^r {}^{13}C_r$.

The summation encompasses the odd-positioned terms of the binomial expansion of $(1+x)^{13}$ with $x=2$, specifically excluding the $r=1$ term. Recalling the identity $(1+x)^{13} - (1-x)^{13} = 2 \sum_{r \text{ odd}} {}^{13}C_r x^r$, we determine that the sum of all odd terms is $\frac{3^{13} - (-1)^{13}}{2} = \frac{3^{13} + 1}{2}$. To isolate our specific series, we subtract the missing $r=1$ term, which is ${}^{13}C_1 \cdot 2^1 = 26$, resulting in $\frac{3^{13} + 1}{2} - 26$. Multiplying this entire value by the external factor of $2$ yields $3^{13} + 1 - 52 = 3^{13} - 51$, identifying $\alpha$ as $51$.

To determine the highest power of a prime that divides a factorial, Legendre's Formula is the standard approach, which involves summing the floor of the number divided by increasing powers of the prime. For the prime 3 and the factorial 50!, we first calculate $\lfloor \frac{50}{3} \rfloor$, which is 16. We then divide 50 by $3^{2}$ to find $\lfloor \frac{50}{9} \rfloor$, which is 5. Continuing this process, we divide 50 by $3^{3}$ to find $\lfloor \frac{50}{27} \rfloor$, which is 1, and since dividing 50 by $3^{4}$ results in a value less than 1, we stop there. Summing these quotients gives $16 + 5 + 1 = 22$, which is the largest integer $n$ such that $3^{n}$ divides 50!.

To evaluate $\alpha$, recognize that the sum $\sum_{r=1}^{6}(-3)^{r-1}\;{}^{12}C_{2r-1}$ extracts the odd-indexed terms from the binomial expansion of $(1+\sqrt{3}i)^{12}$ when scaled by the appropriate complex factor. Using the binomial theorem, the sum of terms with odd powers of $x$ in the expansion of $(1+x)^{12}$ is represented by the formula $\frac{(1+x)^{12} - (1-x)^{12}}{2}$. By setting $x=\sqrt{3}i$, we note that $1+\sqrt{3}i$ can be written in polar form as $2(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3})$. Raising this complex number to the 12th power yields $2^{12}(\cos 4\pi + i\sin 4\pi)$, which simplifies to 4096. Because the conjugate $1-\sqrt{3}i$ raised to the 12th power also results in 4096, the difference between these two quantities is 0, confirming that $\alpha = 1 + 0 = 1$.

With the line equation simplified to $x - \sqrt{3}y + 1 = 0$, the perpendicular distance from the point $(12, \sqrt{3})$ is determined using the formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$. Substituting the given coordinates and line coefficients $A=1$, $B=-\sqrt{3}$, and $C=1$ into this expression gives $d = \frac{|1(12) - \sqrt{3}(\sqrt{3}) + 1|}{\sqrt{1^2 + (-\sqrt{3})^2}}$. Simplifying the numerator leads to $|12 - 3 + 1| = 10$, while the denominator becomes $\sqrt{1 + 3} = 2$. Dividing these values provides the final distance of 5.

The expansion of algebraic expressions involving binomial powers relies on the standard binomial series expansion where $(1+x)^n = 1 + nx + \frac{n(n-1)}{2}x^2 + \dots$. By expressing both terms in the given product as their respective series and focusing on terms up to $x^2$, we identify the coefficients by grouping terms that multiply to produce $x$ or $x^2$.

Specifically, $(1+x)^p = 1 + px + \frac{p(p-1)}{2}x^2$ and $(1-x)^q = 1 - qx + \frac{q(q-1)}{2}x^2$, which multiply to give a linear coefficient of $p-q$ and a quadratic coefficient of $\frac{q(q-1)}{2} + \frac{p(p-1)}{2} - pq$. Given these coefficients are 1 and -2, we establish the system $p-q=1$ and $\frac{q(q-1)}{2} + \frac{p(p-1)}{2} - pq = -2$. Using the identity $(p-q)^2 = p^2 - 2pq + q^2$, the second equation simplifies to $\frac{(p-q)^2 - (p+q)}{2} = -2$. Substituting $p-q=1$ yields $1 - (p+q) = -4$, or $p+q=5$. Solving the system $p-q=1$ and $p+q=5$ provides $p=3$ and $q=2$, resulting in $p^2+q^2=13$.