Binomial Theorem — Class 11 Maths Notes

Topics covered in Binomial Theorem

1. 1.Introduction of Binomial Theorem

   What is the Binomial Theorem?

   The Binomial Theorem provides a way to expand expressions that are raised to a power. It's a very important concept in algebra and has applications in various fields like statistics, probability, and even in everyday calculations.

   The Basic Idea

   If you have an expression like (𝑎+𝑏)𝑛(a+b)n, the Binomial Theorem tells you how to expand it without multiplying the expression by itself 𝑛n times.

   The Binomial Theorem Formula

   The Binomial Theorem states:

   (𝑎+𝑏)𝑛=∑𝑘=0𝑛(𝑛𝑘)𝑎𝑛−𝑘𝑏𝑘(a+b)n=k=0∑n​(kn​)an−kbk

   Here, (𝑛𝑘)(kn​) (read as "n choose k") represents the binomial coefficients, which are calculated as:

   (𝑛𝑘)=𝑛!𝑘!(𝑛−𝑘)!(kn​)=k!(n−k)!n!​

   Step-by-Step Explanation

   1. Understand the Expression: Let's consider (𝑎+𝑏)3(a+b)3. According to the theorem, we can expand this expression without multiplying (𝑎+𝑏)(a+b) by itself three times.

   2. Identify the Power 𝑛n: In this case, 𝑛=3n=3.

   3. Apply the Formula: Use the binomial theorem formula to expand the expression.

   Example: Expanding (𝑎+𝑏)3(a+b)3

   1. Write the general formula: (𝑎+𝑏)𝑛=∑𝑘=0𝑛(𝑛𝑘)𝑎𝑛−𝑘𝑏𝑘(a+b)n=∑k=0n​(kn​)an−kbk

   2. Substitute 𝑛=3n=3: (𝑎+𝑏)3=∑𝑘=03(3𝑘)𝑎3−𝑘𝑏𝑘(a+b)3=∑k=03​(k3​)a3−kbk

   3. Calculate each term:

      • For 𝑘=0k=0: (30)𝑎3𝑏0=1⋅𝑎3=𝑎3(03​)a3b0=1⋅a3=a3
      • For 𝑘=1k=1: (31)𝑎2𝑏1=3⋅𝑎2𝑏=3𝑎2𝑏(13​)a2b1=3⋅a2b=3a2b
      • For 𝑘=2k=2: (32)𝑎1𝑏2=3⋅𝑎𝑏2=3𝑎𝑏2(23​)a1b2=3⋅ab2=3ab2
      • For 𝑘=3k=3: (33)𝑎0𝑏3=1⋅𝑏3=𝑏3(33​)a0b3=1⋅b3=b3

   4. Combine all terms:

      (𝑎+𝑏)3=𝑎3+3𝑎2𝑏+3𝑎𝑏2+𝑏3(a+b)3=a3+3a2b+3ab2+b3

   Real-Life Example

   Suppose you want to calculate the possible outcomes of flipping a coin 3 times. You can use the binomial theorem to find the probabilities of different combinations of heads (H) and tails (T). Each flip has two outcomes (H or T), so you can represent the problem using (𝐻+𝑇)3(H+T)3 and expand it using the binomial theorem.

   Activities to Understand Better

   1. Practice Expanding: Use the binomial theorem to expand (𝑥+𝑦)4(x+y)4 and (2+3)5(2+3)5.
   2. Binomial Coefficients: Calculate binomial coefficients for different values of 𝑛n and 𝑘k using (𝑛𝑘)=𝑛!𝑘!(𝑛−𝑘)!(kn​)=k!(n−k)!n!​.
   3. Pascal's Triangle: Draw Pascal's triangle and use it to find binomial coefficients for expansions.

   Careers and Real-Life Applications

   • Statistics and Probability: The binomial theorem is used in calculating probabilities and in the binomial distribution.
   • Engineering: Engineers use the binomial theorem in various calculations, especially in dealing with polynomials.
   • Finance: Financial analysts use it for modeling and predicting various financial outcomes.
   • Computer Science: Algorithms involving combinatorial calculations often use the binomial theorem.

2. 2.Binomial Theorem for Positive Integral Indices

   What is it?
   

   The Binomial Theorem for positive integral indices provides a way to expand expressions of the form (𝑎+𝑏)𝑛(a+b)n, where 𝑛n is a positive integer. This theorem is crucial in algebra and has applications in various fields, including statistics, probability, and computer science.

   The Binomial Theorem Formula

   For any positive integer 𝑛n:

   (𝑎+𝑏)𝑛=∑𝑘=0𝑛(𝑛𝑘)𝑎𝑛−𝑘𝑏𝑘(a+b)n=k=0∑n​(kn​)an−kbk

   Where:

   • (𝑛𝑘)(kn​) (read as "n choose k") is the binomial coefficient, calculated as:

   (𝑛𝑘)=𝑛!𝑘!(𝑛−𝑘)!(kn​)=k!(n−k)!n!​

   • 𝑎a and 𝑏b are any numbers (they can be variables, constants, or more complex expressions).
   • The sum runs from 𝑘=0k=0 to 𝑘=𝑛k=n.

   Understanding the Components

   1. Binomial Coefficient: (𝑛𝑘)(kn​) represents the number of ways to choose 𝑘k elements from 𝑛n elements and is given by the formula:

   (𝑛𝑘)=𝑛!𝑘!(𝑛−𝑘)!(kn​)=k!(n−k)!n!​

   1. Terms of the Expansion: Each term in the expansion has the form (𝑛𝑘)𝑎𝑛−𝑘𝑏𝑘(kn​)an−kbk.

   Step-by-Step Example: Expanding (𝑎+𝑏)4(a+b)4

   1. Write the general formula:

      (𝑎+𝑏)𝑛=∑𝑘=0𝑛(𝑛𝑘)𝑎𝑛−𝑘𝑏𝑘(a+b)n=k=0∑n​(kn​)an−kbk

   2. Substitute 𝑛=4n=4:

      (𝑎+𝑏)4=∑𝑘=04(4𝑘)𝑎4−𝑘𝑏𝑘(a+b)4=k=0∑4​(k4​)a4−kbk

   3. Calculate each term:

      • For 𝑘=0k=0: (40)𝑎4𝑏0=1⋅𝑎4=𝑎4(04​)a4b0=1⋅a4=a4
      • For 𝑘=1k=1: (41)𝑎3𝑏1=4⋅𝑎3𝑏=4𝑎3𝑏(14​)a3b1=4⋅a3b=4a3b
      • For 𝑘=2k=2: (42)𝑎2𝑏2=6⋅𝑎2𝑏2=6𝑎2𝑏2(24​)a2b2=6⋅a2b2=6a2b2
      • For 𝑘=3k=3: (43)𝑎1𝑏3=4⋅𝑎𝑏3=4𝑎𝑏3(34​)a1b3=4⋅ab3=4ab3
      • For 𝑘=4k=4: (44)𝑎0𝑏4=1⋅𝑏4=𝑏4(44​)a0b4=1⋅b4=b4

   4. Combine all terms:

      (𝑎+𝑏)4=𝑎4+4𝑎3𝑏+6𝑎2𝑏2+4𝑎𝑏3+𝑏4(a+b)4=a4+4a3b+6a2b2+4ab3+b4

   Real-Life Application: Probability

   Let's consider the probability of flipping a coin 4 times and getting heads (H) or tails (T). Using the binomial theorem, you can determine the number of ways to get different combinations of heads and tails.

   For example, the expansion of (𝐻+𝑇)4(H+T)4 gives us:

   (𝐻+𝑇)4=𝐻4+4𝐻3𝑇+6𝐻2𝑇2+4𝐻𝑇3+𝑇4(H+T)4=H4+4H3T+6H2T2+4HT3+T4

   This expansion shows the different combinations of heads and tails when flipping the coin 4 times.

   Practice Problems

   1. Expand (𝑥+𝑦)5(x+y)5 using the binomial theorem.
   2. Find the coefficient of 𝑥3𝑦2x3y2 in the expansion of (𝑥+𝑦)5(x+y)5.

   Careers and Real-Life Applications

   • Statistics and Probability: Used for calculating probabilities and in statistical models.
   • Engineering: Used in polynomial approximations and in various engineering calculations.
   • Finance: Helps in modeling financial scenarios and predicting outcomes.
   • Computer Science: Algorithms involving combinatorial calculations often use the binomial theorem.

3. 3.Pascal’s Triangle

   What is Pascal’s Triangle?

   Pascal’s Triangle is a triangular array of numbers, where each number is the sum of the two numbers directly above it. It has numerous applications in algebra, combinatorics, and probability theory.

   Construction of Pascal’s Triangle

   1. Start with the number 1 at the top (row 0).
   2. Each row starts and ends with 1.
   3. Every other number is the sum of the two numbers directly above it from the previous row.

   Here is the beginning of Pascal’s Triangle:

   Understanding Pascal’s Triangle

   1. Row Number: The first row (with just 1) is considered row 0.
   2. Binomial Coefficients: The numbers in each row are the binomial coefficients (𝑛𝑘)(kn​) which appear in the binomial expansion of (𝑎+𝑏)𝑛(a+b)n.

   Properties of Pascal’s Triangle

   1. Symmetry: Each row is symmetric.
   2. Binomial Coefficients: The 𝑘k-th entry in the 𝑛n-th row represents (𝑛𝑘)(kn​).
   3. Sum of Elements: The sum of the elements in the 𝑛n-th row is 2𝑛2n.
   4. Fibonacci Sequence: The sums of the shallow diagonals give the Fibonacci sequence.

   How to Use Pascal’s Triangle for Binomial Expansion

   To expand (𝑎+𝑏)𝑛(a+b)n using Pascal’s Triangle, use the 𝑛n-th row of the triangle:

   (𝑎+𝑏)𝑛=∑𝑘=0𝑛(𝑛𝑘)𝑎𝑛−𝑘𝑏𝑘(a+b)n=k=0∑n​(kn​)an−kbk

   For example, to expand (𝑎+𝑏)4(a+b)4:

   1. Look at the 4th row of Pascal’s Triangle: 1,4,6,4,11,4,6,4,1.
   2. These are the coefficients for each term in the expansion.
   3. Write the expansion:(𝑎+𝑏)4=1𝑎4+4𝑎3𝑏+6𝑎2𝑏2+4𝑎𝑏3+1𝑏4(a+b)4=1a4+4a3b+6a2b2+4ab3+1b4

   Real-Life Example

   Imagine you are organizing a small event and you have four types of tasks (a, b, c, d) that need to be done. Using Pascal’s Triangle, you can figure out how many ways you can distribute these tasks if each person must do at least one task. For example, the combinations for distributing four tasks can be represented using the binomial coefficients from the 4th row of Pascal’s Triangle.

   Activities

   1. Construct Pascal’s Triangle: Write out the first 6 rows of Pascal’s Triangle.
   2. Find Binomial Coefficients: Use Pascal’s Triangle to find (52)(25​) and (63)(36​).
   3. Expand Using Pascal’s Triangle: Expand (𝑥+𝑦)5(x+y)5 using the 5th row of Pascal’s Triangle.

   Careers and Real-Life Applications

   • Combinatorics: Pascal’s Triangle is essential in combinatorial problems and counting methods.
   • Probability: It helps in calculating probabilities in binomial distributions.
   • Algebra: Useful in expanding binomials and solving polynomial equations.
   • Computer Science: Algorithms involving combinations and permutations often use concepts from Pascal’s Triangle.

4. 4.Binomial Theorem for Any Positive Integer 𝑛

   What is it?

   The Binomial Theorem for any positive integer 𝑛n provides a way to expand expressions of the form (𝑎+𝑏)𝑛(a+b)n. This theorem is a fundamental concept in algebra and has applications in various fields such as combinatorics, probability, and calculus.

   The Binomial Theorem Formula

   For any positive integer 𝑛n:

   (𝑎+𝑏)𝑛=∑𝑘=0𝑛(𝑛𝑘)𝑎𝑛−𝑘𝑏𝑘(a+b)n=k=0∑n​(kn​)an−kbk

   Where:

   • (𝑛𝑘)(kn​) (read as "n choose k") is the binomial coefficient, calculated as:

   (𝑛𝑘)=𝑛!𝑘!(𝑛−𝑘)!(kn​)=k!(n−k)!n!​

   • 𝑎a and 𝑏b are any numbers (they can be variables, constants, or more complex expressions).
   • The sum runs from 𝑘=0k=0 to 𝑘=𝑛k=n.

   Understanding the Components

   1. Binomial Coefficient: (𝑛𝑘)(kn​) represents the number of ways to choose 𝑘k elements from 𝑛n elements and is given by the formula:

   (𝑛𝑘)=𝑛!𝑘!(𝑛−𝑘)!(kn​)=k!(n−k)!n!​

   1. Terms of the Expansion: Each term in the expansion has the form (𝑛𝑘)𝑎𝑛−𝑘𝑏𝑘(kn​)an−kbk.

   Step-by-Step Example: Expanding (𝑎+𝑏)5(a+b)5

   1. Write the general formula:

      (𝑎+𝑏)𝑛=∑𝑘=0𝑛(𝑛𝑘)𝑎𝑛−𝑘𝑏𝑘(a+b)n=k=0∑n​(kn​)an−kbk

   2. Substitute 𝑛=5n=5:

      (𝑎+𝑏)5=∑𝑘=05(5𝑘)𝑎5−𝑘𝑏𝑘(a+b)5=k=0∑5​(k5​)a5−kbk

   3. Calculate each term:

      • For 𝑘=0k=0: (50)𝑎5𝑏0=1⋅𝑎5=𝑎5(05​)a5b0=1⋅a5=a5
      • For 𝑘=1k=1: (51)𝑎4𝑏1=5⋅𝑎4𝑏=5𝑎4𝑏(15​)a4b1=5⋅a4b=5a4b
      • For 𝑘=2k=2: (52)𝑎3𝑏2=10⋅𝑎3𝑏2=10𝑎3𝑏2(25​)a3b2=10⋅a3b2=10a3b2
      • For 𝑘=3k=3: (53)𝑎2𝑏3=10⋅𝑎2𝑏3=10𝑎2𝑏3(35​)a2b3=10⋅a2b3=10a2b3
      • For 𝑘=4k=4: (54)𝑎1𝑏4=5⋅𝑎𝑏4=5𝑎𝑏4(45​)a1b4=5⋅ab4=5ab4
      • For 𝑘=5k=5: (55)𝑎0𝑏5=1⋅𝑏5=𝑏5(55​)a0b5=1⋅b5=b5

   4. Combine all terms:

      (𝑎+𝑏)5=𝑎5+5𝑎4𝑏+10𝑎3𝑏2+10𝑎2𝑏3+5𝑎𝑏4+𝑏5(a+b)5=a5+5a4b+10a3b2+10a2b3+5ab4+b5

   Real-Life Application: Combinations

   The binomial theorem is closely related to combinations, which are used in various real-life scenarios such as determining possible outcomes in games, distributing tasks, or forming committees. For example, if you want to find out how many ways you can form a committee of 3 members from a group of 5 people, you would use the binomial coefficient (53)(35​).

   Practice Problems

   1. Expand (𝑥+𝑦)6(x+y)6 using the binomial theorem.
   2. Find the coefficient of 𝑥4𝑦x4y in the expansion of (𝑥+𝑦)5(x+y)5.
   3. Calculate the binomial coefficient (73)(37​) and use it to expand (𝑎+𝑏)7(a+b)7.

   Careers and Real-Life Applications

   • Statistics and Probability: Used for calculating probabilities and in binomial distributions.
   • Engineering: Helps in polynomial approximations and various engineering calculations.
   • Finance: Assists in modeling financial scenarios and predicting outcomes.
   • Computer Science: Algorithms involving combinatorial calculations often use the binomial theorem.

5. 5.Exercise Questions

   Question:

   Using the binomial theorem, prove that 6𝑛−5𝑛6n−5n always leaves a remainder of 1 when divided by 25.

   Solution:

   To prove that 6𝑛−5𝑛6n−5n leaves a remainder of 1 when divided by 25, we need to show that 6𝑛−5𝑛≡1 (mod 25)6n−5n≡1 (mod 25).

   We start by using the binomial theorem to express 6𝑛6n. Notice that 6≡1 (mod 5)6≡1 (mod 5), and thus:

   6𝑛≡1𝑛≡1 (mod 5)6n≡1n≡1 (mod 5)

   This suggests that 6𝑛−16n−1 is divisible by 5, i.e., 6𝑛≡1 (mod 5)6n≡1 (mod 5).

   Now, let's consider the expression modulo 25. We will use the binomial expansion:

   (5+1)𝑛=∑𝑘=0𝑛(𝑛𝑘)5𝑘⋅1𝑛−𝑘(5+1)n=k=0∑n​(kn​)5k⋅1n−k

   This expands to:

   6𝑛=(5+1)𝑛=∑𝑘=0𝑛(𝑛𝑘)5𝑘6n=(5+1)n=k=0∑n​(kn​)5k

   Separating the terms, we have:

   6𝑛=1𝑛+(𝑛1)5+(𝑛2)52+⋯+(𝑛𝑛)5𝑛6n=1n+(1n​)5+(2n​)52+⋯+(nn​)5n

   This simplifies to:

   6𝑛=1+𝑛⋅5+(𝑛2)52+⋯+(𝑛𝑛)5𝑛6n=1+n⋅5+(2n​)52+⋯+(nn​)5n

   Now, let's express 6𝑛6n modulo 25:

   6𝑛=1+5𝑛+52(𝑛2)+53(𝑛3)+⋯6n=1+5n+52(2n​)+53(3n​)+⋯

   When taken modulo 25, all terms containing 5252 or higher powers of 5 will be 0, because 52=2552=25 and higher powers are multiples of 25. Hence, these terms vanish in modulo 25:

   6𝑛≡1+5𝑛 (mod 25)6n≡1+5n (mod 25)

   Thus, we have:

   6𝑛−5𝑛≡1+5𝑛−5𝑛 (mod 25)6n−5n≡1+5n−5n (mod 25)

   Simplifying:

   6𝑛−5𝑛≡1 (mod 25)6n−5n≡1 (mod 25)

   Therefore, 6𝑛−5𝑛6n−5n always leaves a remainder of 1 when divided by 25.

6. 6.Using the binomial theorem, evaluate each of the following:

   1. (96)3(96)3
   2. (102)5(102)5
   3. (101)4(101)4
   4. Using Binomial Theorem, indicate which number is larger: (1.1)10000(1.1)10000 or 1000.
   5. Find (𝑎+𝑏)4−(𝑎−𝑏)4(a+b)4−(a−b)4. Hence, evaluate (3+2)4−(3−2)4(3​+2​)4−(3​−2​)4.
   6. Find (𝑥+1)6+(𝑥−1)6(x+1)6+(x−1)6. Hence or otherwise evaluate (2+1)6+(2−1)6(2​+1)6+(2​−1)6.
   7. Show that 9𝑛+1−8𝑛−99n+1−8n−9 is divisible by 64, whenever 𝑛n is a positive integer.
   8. Prove that ∑𝑟=0𝑛3𝑟(𝑛𝑟)=4𝑛∑r=0n​3r(rn​)=4n.

   ---

   Solutions:

   1. Evaluate (96)3(96)3 using the binomial theorem:

   96=100−496=100−4(96)3=(100−4)3(96)3=(100−4)3

   Using the binomial theorem:

   (𝑎−𝑏)3=𝑎3−3𝑎2𝑏+3𝑎𝑏2−𝑏3(a−b)3=a3−3a2b+3ab2−b3

   Here, 𝑎=100a=100 and 𝑏=4b=4:

   (100−4)3=1003−3⋅1002⋅4+3⋅100⋅42−43(100−4)3=1003−3⋅1002⋅4+3⋅100⋅42−43=1000000−3⋅10000⋅4+3⋅100⋅16−64=1000000−3⋅10000⋅4+3⋅100⋅16−64=1000000−120000+4800−64=1000000−120000+4800−64=1000000−120000+4800−64=884736=1000000−120000+4800−64=884736

   So, (96)3=884736(96)3=884736.

   2. Evaluate (102)5(102)5 using the binomial theorem:

   102=100+2102=100+2(102)5=(100+2)5(102)5=(100+2)5

   Using the binomial theorem:

   (𝑎+𝑏)5=𝑎5+5𝑎4𝑏+10𝑎3𝑏2+10𝑎2𝑏3+5𝑎𝑏4+𝑏5(a+b)5=a5+5a4b+10a3b2+10a2b3+5ab4+b5

   Here, 𝑎=100a=100 and 𝑏=2b=2:

   (100+2)5=1005+5⋅1004⋅2+10⋅1003⋅22+10⋅1002⋅23+5⋅100⋅24+25(100+2)5=1005+5⋅1004⋅2+10⋅1003⋅22+10⋅1002⋅23+5⋅100⋅24+25=10000000000+5⋅100000000⋅2+10⋅1000000⋅4+10⋅10000⋅8+5⋅100⋅16+32=10000000000+5⋅100000000⋅2+10⋅1000000⋅4+10⋅10000⋅8+5⋅100⋅16+32=10000000000+1000000000+40000000+800000+8000+32=10000000000+1000000000+40000000+800000+8000+32=10000000000+1000000000+40000000+800000+8000+32=11120808032=10000000000+1000000000+40000000+800000+8000+32=11120808032

   So, (102)5=11120808032(102)5=11120808032.

   3. Evaluate (101)4(101)4 using the binomial theorem:

   101=100+1101=100+1(101)4=(100+1)4(101)4=(100+1)4

   Using the binomial theorem:

   (𝑎+𝑏)4=𝑎4+4𝑎3𝑏+6𝑎2𝑏2+4𝑎𝑏3+𝑏4(a+b)4=a4+4a3b+6a2b2+4ab3+b4

   Here, 𝑎=100a=100 and 𝑏=1b=1:

   (100+1)4=1004+4⋅1003⋅1+6⋅1002⋅12+4⋅100⋅13+14(100+1)4=1004+4⋅1003⋅1+6⋅1002⋅12+4⋅100⋅13+14=100000000+4⋅1000000+6⋅10000+4⋅100+1=100000000+4⋅1000000+6⋅10000+4⋅100+1=100000000+4000000+60000+400+1=100000000+4000000+60000+400+1=104060401=104060401

   So, (101)4=104060401(101)4=104060401.

   4. Using Binomial Theorem, indicate which number is larger: (1.1)10000(1.1)10000 or 1000.

   First, note that:

   1.1=1+0.11.1=1+0.1

   Using the binomial theorem for large 𝑛n:

   (1+𝑥)𝑛≈𝑒𝑛𝑥(1+x)n≈enx

   Here, 𝑥=0.1x=0.1 and 𝑛=10000n=10000:

   (1.1)10000≈𝑒10000⋅0.1=𝑒1000(1.1)10000≈e10000⋅0.1=e1000

   Since 𝑒≈2.718e≈2.718:

   𝑒1000 is much larger than 1000e1000 is much larger than 1000

   Thus, (1.1)10000(1.1)10000 is much larger than 1000.

   5. Find (𝑎+𝑏)4−(𝑎−𝑏)4(a+b)4−(a−b)4. Hence, evaluate (3+2)4−(3−2)4(3​+2​)4−(3​−2​)4.

   Using the binomial theorem:

   (𝑎+𝑏)4=𝑎4+4𝑎3𝑏+6𝑎2𝑏2+4𝑎𝑏3+𝑏4(a+b)4=a4+4a3b+6a2b2+4ab3+b4(𝑎−𝑏)4=𝑎4−4𝑎3𝑏+6𝑎2𝑏2−4𝑎𝑏3+𝑏4(a−b)4=a4−4a3b+6a2b2−4ab3+b4

   Subtracting:

   (𝑎+𝑏)4−(𝑎−𝑏)4=(𝑎4+4𝑎3𝑏+6𝑎2𝑏2+4𝑎𝑏3+𝑏4)−(𝑎4−4𝑎3𝑏+6𝑎2𝑏2−4𝑎𝑏3+𝑏4)(a+b)4−(a−b)4=(a4+4a3b+6a2b2+4ab3+b4)−(a4−4a3b+6a2b2−4ab3+b4)=8𝑎3𝑏+8𝑎𝑏3=8a3b+8ab3=8𝑎𝑏(𝑎2+𝑏2)=8ab(a2+b2)

   For 𝑎=3a=3​ and 𝑏=2b=2​:

   (3+2)4−(3−2)4=832(32+22)(3​+2​)4−(3​−2​)4=83​2​(3​2+2​2)=8⋅6⋅(3+2)=8⋅6​⋅(3+2)=8⋅6⋅5=8⋅6​⋅5=406=406​

   So, (3+2)4−(3−2)4=406(3​+2​)4−(3​−2​)4=406​.

   6. Find (𝑥+1)6+(𝑥−1)6(x+1)6+(x−1)6. Hence or otherwise evaluate (2+1)6+(2−1)6(2​+1)6+(2​−1)6.

   Using the binomial theorem:

   (𝑥+1)6=𝑥6+6𝑥5+15𝑥4+20𝑥3+15𝑥2+6𝑥+1(x+1)6=x6+6x5+15x4+20x3+15x2+6x+1(𝑥−1)6=𝑥6−6𝑥5+15𝑥4−20𝑥3+15𝑥2−6𝑥+1(x−1)6=x6−6x5+15x4−20x3+15x2−6x+1

   Adding:

   (𝑥+1)6+(𝑥−1)6=(𝑥6+6𝑥5+15𝑥4+20𝑥3+15𝑥2+6𝑥+1)+(𝑥6−6𝑥5+15𝑥4−20𝑥3+15𝑥2−6𝑥+1)(x+1)6+(x−1)6=(x6+6x5+15x4+20x3+15x2+6x+1)+(x6−6x5+15x4−20x3+15x2−6x+1)=2𝑥6+30𝑥4+30𝑥2+2=2x6+30x4+30x2+2

   For 𝑥=2x=2​:

   (2+1)6+(2−1)6=2(2)6+30(2)4+30(2)2+2(2​+1)6+(2​−1)6=2(2​)6+30(2​)4+30(2​)2+2=2⋅8+30⋅4+30⋅2+2=2⋅8+30⋅4+30⋅2+2=16+120+60+2=16+120+60+2=198=198

   So, (2+1)6+(2−1)6=198(2​+1)6+(2​−1)6=198.

   7. Show that 9𝑛+1−8𝑛−99n+1−8n−9 is divisible by 64, whenever 𝑛n is a positive integer.

   Consider the expression 9𝑛+1−8𝑛−99n+1−8n−9:

   Using 9=8+19=8+1:

   9𝑛+1=(8+1)𝑛+19n+1=(8+1)n+1

   Using the binomial theorem:

   (8+1)𝑛+1=∑𝑘=0𝑛+1(𝑛+1𝑘)8𝑘(8+1)n+1=k=0∑n+1​(kn+1​)8k=1+(𝑛+1)8+(𝑛+1)𝑛282+⋯+8𝑛+1=1+(n+1)8+2(n+1)n​82+⋯+8n+1

   Now, consider the expression modulo 64:

   9𝑛+1≡(1+(𝑛+1)8+(𝑛+1)𝑛282+⋯+8𝑛+1) (mod 64)9n+1≡(1+(n+1)8+2(n+1)n​82+⋯+8n+1) (mod 64)

   Only the first two terms are relevant:

   1+(𝑛+1)8 (mod 64)1+(n+1)8 (mod 64)

   Subtracting 8𝑛+98n+9:

   9𝑛+1−8𝑛−9≡(1+(𝑛+1)8)−8𝑛−9 (mod 64)9n+1−8n−9≡(1+(n+1)8)−8n−9 (mod 64)

   Simplifying:

   =1+8𝑛+8−8𝑛−9=1+8n+8−8n−9=0 (mod 64)=0 (mod 64)

   So, 9𝑛+1−8𝑛−99n+1−8n−9 is divisible by 64.

   8. Prove that ∑𝑟=0𝑛3𝑟(𝑛𝑟)=4𝑛∑r=0n​3r(rn​)=4n.

   Consider the binomial expansion of (1+3)𝑛(1+3)n:

   (1+3)𝑛=∑𝑟=0𝑛(𝑛𝑟)1𝑛−𝑟3𝑟(1+3)n=r=0∑n​(rn​)1n−r3r4𝑛=∑𝑟=0𝑛(𝑛𝑟)3𝑟4n=r=0∑n​(rn​)3r

   So, ∑𝑟=0𝑛3𝑟(𝑛𝑟)=4𝑛∑r=0n​3r(rn​)=4n.

   Thus, the equation is proved.