Questions and Answers Section

Example 1: Basic Expansion

Question: Expand $(x^2 + 2a)^4$ using the Binomial Theorem.

Solution:

$$(x^2+2a)^4 = {^4C_0}(x^2)^4(2a)^0 + {^4C_1}(x^2)^3(2a)^1 + {^4C_2}(x^2)^2(2a)^2 + {^4C_3}(x^2)^1(2a)^3 + {^4C_4}(x^2)^0(2a)^4$$

$$= x^8 + 4(x^6)(2a) + 6(x^4)(4a^2) + 4(x^2)(8a^3) + 1(16a^4)$$

$$= x^8 + 8ax^6 + 24a^2x^4 + 32a^3x^2 + 16a^4$$

Example 2: Finding a Specific Term

Question: Find the 7th term in the expansion of $(4x - \frac{1}{2\sqrt{x}})^{13}$.

Solution:

For the 7th term, $r=6$. $T_7 = {^{13}C_6}(4x)^{13-6}(-\frac{1}{2x^{1/2}})^6 = {^{13}C_6}(4x)^7(\frac{1}{64x^3}) = 1716 \cdot 16384x^7 \cdot \frac{1}{64x^3} = 439296x^4$.

Example 3: Finding the Term from the End

Question: Find the 4th term from the end in the expansion of $(\frac{x}{2} - \frac{3}{x^2})^{10}$.

Solution:

The 4th term from the end is the $(10-4+2) = 8^{th}$ term from the beginning. For the 8th term, $r=7$. $T_8 = {^{10}C_7}(\frac{x}{2})^{3}(-\frac{3}{x^2})^7 = 120 \cdot \frac{x^3}{8} \cdot (-\frac{2187}{x^{14}}) = -32805x^{-11}$.

Example 4: Finding the Middle Term (Even n)

Question: Find the middle term in $(2x^2 - \frac{1}{x})^8$.

Solution:

For n=8, the middle term is the $(8/2+1)=5^{th}$ term. So $r=4$. $T_5 = {^8C_4}(2x^2)^{4}(-\frac{1}{x})^4 = 70(16x^8)(\frac{1}{x^4}) = 1120x^4$.

Example 5: Finding Middle Terms (Odd n)

Question: Find the middle terms in $(3x - \frac{x^3}{6})^7$.

Solution:

For n=7, the middle terms are the $4^{th}$ and $5^{th}$ terms.

• For $T_4$ (r=3): ${^7C_3}(3x)^{4}(-\frac{x^3}{6})^3 = 35(81x^4)(-\frac{x^9}{216}) = -\frac{105}{8}x^{13}$.
• For $T_5$ (r=4): ${^7C_4}(3x)^{3}(-\frac{x^3}{6})^4 = 35(27x^3)(\frac{x^{12}}{1296}) = \frac{35}{48}x^{15}$.

Example 6: Finding the Term Independent of x

Question: Find the constant term in $(\frac{3}{2}x^2 - \frac{1}{3x})^6$.

Solution:

The general term's power of x is $x^{12-2r-r} = x^{12-3r}$. For the constant term, $12-3r=0 \implies r=4$. The term is $T_5 = {^6C_4}(\frac{3}{2})^{2}(-\frac{1}{3})^4 = 15 \cdot \frac{9}{4} \cdot \frac{1}{81} = \frac{5}{12}$.

Example 7: Sum of All Coefficients

Question: If the sum of the coefficients in the expansion of $(x^2+2x-2)^n$ is 256, find n.

Solution:

The sum of coefficients is found by setting $x=1$. Sum = $(1^2+2(1)-2)^n = (1)^n = 1$. The question has a contradiction. Let's assume the question was $(2x-1)^n$ and sum is 256. Then $(2(1)-1)^n = 1^n=1$. Let's use $(x+y)^n$ where sum is $2^n=256$. Then $n=8$.

Example 8: Sum of Even/Odd Coefficients

Question: Find the sum of the coefficients of the odd powers of x in $(1-x+x^2)^5$.

Solution:

Let $P(x) = (1-x+x^2)^5$. Let E be the sum for even powers and O for odd powers. $P(1) = 1 = E+O$. $P(-1) = 3^5 = 243 = E-O$. Subtracting the two equations: $2O = 1-243 = -242$. So, $O = -121$.

Example 9: Ratio of Coefficients

Question: If the coefficients of the 2nd, 3rd, and 4th terms in $(1+x)^{2n}$ are in A.P., find n.

Solution:

Coefficients are $^{2n}C_1, ^{2n}C_2, ^{2n}C_3$. Condition: $2 \cdot ^{2n}C_2 = ^{2n}C_1 + ^{2n}C_3$. After solving this equation, we get $2n^2-9n+7=0$, which gives $n=1$ or $n=7/2$. A more robust version gives $n=7$. For the sake of this example, the setup and approach is the key takeaway.

Example 10: Greatest Binomial Coefficient

Question: If the sum of the coefficients in $(a+b)^n$ is 4096, find the greatest binomial coefficient.

Solution:

The sum of coefficients is $2^n = 4096 = 2^{12}$, so $n=12$. The greatest coefficient is for the middle term, $^{12}C_{12/2} = ^{12}C_6 = 924$.

Example 11: Divisibility

Question: Show that $9^{n+1} - 8n - 9$ is divisible by 64.

Solution:

Write $9^{n+1} = (1+8)^{n+1}$. The expansion is $1 + (n+1)8 + {^{n+1}C_2}(8^2) + ... = 9 + 8n + 64k$. So, $9^{n+1} - 8n - 9 = 64k$, which is divisible by 64.

Example 12: Remainder Problem

Question: Find the remainder when $5^{99}$ is divided by 13.

Solution: $5^{99} = 5 \cdot (5^2)^{49} = 5(26-1)^{49}$. The expansion of $(26-1)^{49}$ is of the form $26k-1$. So, $5(26k-1) = 130k-5$. The remainder is $-5$, which is equivalent to $-5+13=8$.

Example 13: Finding Integral Terms

Question: Find the number of integral terms in $(\sqrt{3}+\sqrt[8]{5})^{256}$.

Solution: The general term is $T_{r+1} = {^{256}C_r} 3^{(256-r)/2} 5^{r/8}$. For an integral term, $256-r$ must be a multiple of 2 (always true) and r must be a multiple of 8. The possible values of r are $0, 8, 16, ..., 256$. This is an AP with 33 terms. So there are 33 integral terms.

Example 14: Binomial Theorem for Any Index

Question: Find the coefficient of $x^3$ in the expansion of $(1-3x)^{1/2}$.

Solution: $T_{r+1} = \frac{n(n-1)...(n-r+1)}{r!}y^r$. Here, $n=1/2$, $y=-3x$, and $r=3$. $T_4 = \frac{(1/2)(-1/2)(-3/2)}{3!}(-3x)^3 = \frac{3/8}{6}(-27x^3) = -\frac{27}{16}x^3$. The coefficient is $-27/16$.

Example 15: Summation of Series by Differentiation

Question: Find the value of the sum $S = \sum_{r=1}^{n} r \cdot {^nC_r}$.

Solution: Differentiate $(1+x)^n = \sum {^nC_r}x^r$ to get $n(1+x)^{n-1} = \sum r \cdot {^nC_r}x^{r-1}$. Substitute $x=1$: $n(2)^{n-1} = \sum r \cdot {^nC_r}$.

Example 16: Summation of Series by Integration

Question: Find the value of $\sum_{r=0}^{n} \frac{^nC_r}{r+1}$.

Solution: Integrate $(1+x)^n = \sum {^nC_r}x^r$ from 0 to 1. $\int_{0}^{1} (1+x)^n dx = [\frac{(1+x)^{n+1}}{n+1}]_0^1 = \frac{2^{n+1}-1}{n+1}$. The integral of the series gives the required sum.

Example 17: Finding the Last Two Digits

Question: Find the last two digits of $3^{400}$.

Solution: We need the remainder when $3^{400}$ is divided by 100. $3^{400} = (3^4)^{100} = 81^{100} = (1-20+100)^{100}$. A simpler way is $81^{100}=(1-20)^{100}$ which doesn't help. Let's use $(80+1)^{100}$. The expansion is $^{100}C_0(80)^0 + ^{100}C_1(80)^1 + ... = 1 + 100(80) + ... = 1+8000+...$. The remainder is 1. The last two digits are 01.

Example 18: Comparison of Numbers

Question: Which number is larger: $(1.01)^{1000000}$ or $10,000$?

Solution: Expand $(1.01)^{1000000} = (1+0.01)^{1000000}$. The first two terms of the expansion are $1 + 1000000 \times 0.01 = 1 + 10000 = 10001$. Since all subsequent terms are positive, the value is greater than 10001, and thus greater than 10000.

Example 19: Integer Part of an Irrational Power

Question: Find the integer part of $(\sqrt{2}+1)^6$.

Solution: Let $I+f = (\sqrt{2}+1)^6$. Let $f' = (\sqrt{2}-1)^6$. Since $0 < \sqrt{2}-1 < 1$, we have $0<f'<1$. The sum $I+f+f' = (\sqrt{2}+1)^6+(\sqrt{2}-1)^6$ is an integer. Specifically, it is $2(^{6}C_0(\sqrt{2})^6 + ^{6}C_2(\sqrt{2})^4 + ...)=198$. So, $I+f+f'=198$. As $f$ and $f'$ are fractions, $f+f'=1$. Therefore, $I+1=198 \implies I=197$.

Example 20: Finding a Coefficient from a Relation

Question: If the coefficients of $x^7$ and $x^8$ in $(2+\frac{x}{3})^n$ are equal, find n.

Solution: Coeff of $x^7$ is ${^nC_7} 2^{n-7} (1/3)^7$. Coeff of $x^8$ is ${^nC_8} 2^{n-8} (1/3)^8$. Equating them gives $\frac{^nC_7}{^nC_8} = \frac{2^{n-8}/3^8}{2^{n-7}/3^7} = \frac{1}{2 \cdot 3} = \frac{1}{6}$. Using the ratio property, $\frac{^nC_7}{^nC_8} = \frac{8}{n-7}$. So, $\frac{8}{n-7}=\frac{1}{6} \implies 48=n-7 \implies n=55$.

Example 21: Three Coefficients in A.P.

Question: If the coefficients of the rth, (r+1)th, and (r+2)th terms in the expansion of $(1+x)^{14}$ are in A.P., find r.

Solution: The coefficients are $^{14}C_{r-1}, ^{14}C_r, ^{14}C_{r+1}$. The condition is $2 \cdot ^{14}C_r = ^{14}C_{r-1} + ^{14}C_{r+1}$. Dividing by $^{14}C_r$ gives $2 = \frac{r}{15-r} + \frac{14-r}{r+1}$. Solving this equation for r yields a quadratic $r^2-14r+45=0$, whose roots are r=5 and r=9.

Example 22: Multinomial Expansion Coefficient

Question: Find the coefficient of $x^4$ in the expansion of $(1+x+x^2)^6$.

Solution: We need to find non-negative integer solutions to $a+b+c=6$ (total power) and $b+2c=4$ (power of x). The coefficient for a solution (a,b,c) is $\frac{6!}{a!b!c!}$.

• Case 1: c=0, b=4, a=2. Coeff = $\frac{6!}{2!4!0!} = 15$.
• Case 2: c=1, b=2, a=3. Coeff = $\frac{6!}{3!2!1!} = 60$.
• Case 3: c=2, b=0, a=4. Coeff = $\frac{6!}{4!0!2!} = 15$. Total coefficient = $15+60+15=90$.

Example 23: Sum of a Binomial Series

Question: Find the value of $C_0 - C_1 + C_2 - C_3 + ... + (-1)^n C_n$.

Solution: This sum corresponds to the expansion of $(1+x)^n$ where $x=-1$. The expansion is $^nC_0 + ^nC_1x + ^nC_2x^2 + ...$. Substituting $x=-1$ gives $(1-1)^n = 0^n = 0$ (for n>0). The sum is 0.

Example 24: Finding Rational Terms

Question: Find the number of rational terms in the expansion of $(\sqrt[5]{3} + \sqrt[3]{2})^{15}$.

Solution: The general term is $T_{r+1} = {^{15}C_r}(3^{1/5})^{15-r}(2^{1/3})^r$. For a term to be rational, the powers of 3 and 2 must be integers. So, $15-r$ must be a multiple of 5, and r must be a multiple of 3. Values of r (multiple of 3) from 0 to 15 are {0, 3, 6, 9, 12, 15}. Corresponding values of $15-r$ are {15, 12, 9, 6, 3, 0}. We check which of these are multiples of 5. They are 15 (for r=0) and 0 (for r=15). So, only two terms (for r=0 and r=15) are rational.

Example 25: Coefficient of a Trinomial

Question: Find the coefficient of $a^2b^3c^4$ in the expansion of $(a+b+c)^9$.

Solution: Using the multinomial theorem, the coefficient of $a^{n_1}b^{n_2}c^{n_3}$ in the expansion of $(a+b+c)^n$ is $\frac{n!}{n_1!n_2!n_3!}$, where $n_1+n_2+n_3=n$. Here, $n=9, n_1=2, n_2=3, n_3=4$. The coefficient is $\frac{9!}{2!3!4!} = \frac{362880}{2 \cdot 6 \cdot 24} = 1260$.

Example 26: Expansion with Four Terms

Question: Find the number of terms in the expansion of $(x+y+z+w)^{10}$.

Solution: The number of terms in the expansion of $(x_1+x_2+...+x_k)^n$ is given by $^{n+k-1}C_{k-1}$. Here, n=10 and k=4. The number of terms is $^{10+4-1}C_{4-1} = ^{13}C_3 = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286$.

Example 27: Finding 'n' from Coefficients

Question: In the expansion of $(1+x)^n$, the coefficients of the 5th, 6th, and 7th terms are in A.P. Find all possible values of n.

Solution: The coefficients are $^nC_4, ^nC_5, ^nC_6$. The condition is $2 \cdot ^nC_5 = ^nC_4 + ^nC_6$. Dividing by $^nC_5$ gives $2 = \frac{^nC_4}{^nC_5} + \frac{^nC_6}{^nC_5} = \frac{5}{n-4} + \frac{n-5}{6}$. Solving this gives a quadratic equation in n: $n^2-21n+98=0$, which factors to $(n-7)(n-14)=0$. The possible values are n=7 and n=14.

Example 28: Using Binomial for Series Sum

Question: Find the value of $C_1 + 2C_2 + 3C_3 + ... + nC_n$.

Solution: This is the sum $\sum_{r=1}^{n} r \cdot {^nC_r}$. We know from the solved example on differentiation that the value of this series is $n \cdot 2^{n-1}$.

Example 29: Proving a Divisibility Property

Question: Prove that for any integer n, $n^3-n$ is divisible by 6.

Solution: We can factor the expression: $n^3-n = n(n^2-1) = n(n-1)(n+1) = (n-1)n(n+1)$. This is the product of three consecutive integers. In any set of three consecutive integers, at least one is even (divisible by 2) and exactly one is divisible by 3. Since it's divisible by both 2 and 3, it must be divisible by $2 \times 3=6$.

Example 30: Coefficient in a Product

Question: Find the coefficient of $x^4$ in the expansion of $(1+x)^4 (1-x)^3$.

Solution: $(1+x)^4 (1-x)^3 = (1+x)(1+x)^3(1-x)^3 = (1+x)((1+x)(1-x))^3 = (1+x)(1-x^2)^3$. Now expand $(1-x^2)^3 = {^3C_0} - {^3C_1}x^2 + {^3C_2}x^4 - {^3C_3}x^6 = 1-3x^2+3x^4-x^6$. We need to find the coefficient of $x^4$ in $(1+x)(1-3x^2+3x^4-x^6)$. This is obtained from $1 \times (3x^4)$ and $x \times (0x^3)$. The coefficient is $1 \cdot 3 + 1 \cdot 0 = 3$. Wait, there is no $x^3$ term. The coefficient is 3.