RBSE Solutions Class 11 Maths Chapter 7 Binomial Theorem More Questions

Question 1. The number of terms in the expansion of \( \left(\frac{a}{x} + bx\right)^{12} \) are-
(A) 11
(B) 13
(C) 10
(D) 14
Answer: (B) 13
In simple words: When you expand a binomial expression like \( (A + B)^n \), the total number of terms will always be one more than the power 'n'. Here, the power is 12, so there are 13 terms. This is a basic rule for binomial expansions.

🎯 Exam Tip: Remember the formula for the number of terms in a binomial expansion: if the power is n, the number of terms is \( n + 1 \).

Question 3. The middle term in the expansion of \( (a - x)^8 \) is
(A) \( 56a^3 x^5 \)
(B) \( -56a^3 x^5 \)
(C) \( 70a^4 x^4 \)
(D) \( -70a^4 x^4 \)
Answer: (C) \( 70a^4 x^4 \)
In simple words: For a binomial expansion with an even power, the middle term is found by taking the power, dividing it by two, and adding one. Then, you calculate that specific term using the binomial theorem. In this case, the middle term is the 5th term, which simplifies to \( 70a^4 x^4 \).

🎯 Exam Tip: When the power 'n' is even, the middle term is the \( \left(\frac{n}{2} + 1\right)^{th} \) term. Pay close attention to the signs in the expansion, especially when there's a minus sign in the binomial.

Question 4. The constant term in the expansion \( \left(2x+\frac{1}{3x^2}\right)^9 \) is
(A) 5th
(B) 4th
(C) 6th
(D) 7th
Answer: (C) 6th
In simple words: The constant term is the one that does not have 'x' in it, meaning the power of 'x' is zero. By finding the value of 'r' for which the power of 'x' becomes zero, we identify that the 6th term is the constant term in this expansion.

🎯 Exam Tip: To find the constant term, set the power of x in the general term \( T_{r+1} \) to zero and solve for r. Remember that \( x^a \cdot x^b = x^{a+b} \) and \( \frac{x^a}{x^b} = x^{a-b} \).

Question 5. The general term in the expansion of \( (x + a)^n \) is
(A) \( {}^nC_r x^{n-r} a^r \)
(B) \( {}^nC_r x a^r \)
(C) \( {}^{n}C_{n-r} x^{n-r} a \)
(D) \( {}^{n}C_{n-r} x a^{n-r} \)
Answer: (A) \( {}^nC_r x^{n-r} a^r \)
In simple words: The general term, also called the \( (r+1)^{th} \) term, tells us the formula for any term in the binomial expansion of \( (x + a)^n \). It combines combinations with powers of x and a. This formula is fundamental for working with binomial theorems.

🎯 Exam Tip: Memorize the general term formula \( T_{r+1} = {}^nC_r x^{n-r} a^r \). This formula is key for finding specific terms, coefficients, or terms independent of x.

Question 6. The value of term independent of x in the expansion \( \left(2x^2-\frac{1}{x}\right)^{12} \) is
(A) 264
(B) -264
(C) 7920
(D) -7920
Answer: (C) 7920
In simple words: To find the term that does not contain 'x', we first write the general term of the expansion. Then, we set the total power of 'x' in this general term to zero. This helps us find the value of 'r', which tells us the position of the constant term. Once 'r' is known, we substitute it back into the general term to get the numerical value.

🎯 Exam Tip: Be careful with the exponents and signs when simplifying the powers of x, especially when x is in the denominator. Remember that \( \frac{1}{x} = x^{-1} \) and \( (x^a)^b = x^{ab} \).

Question 7. The coefficient of \( x^{-17} \) in the expansion of \( \left(x^4-\frac{1}{x^3}\right)^{15} \) is
(A) 1365
(B) -1365
(C) 3003
(D) -3003
Answer: (B) -1365
In simple words: To find the coefficient of a specific power of x, first write out the general term of the binomial expansion. Next, make the power of x in the general term equal to the target power, which is \( -17 \) here. Solve for 'r', then substitute this 'r' value back into the coefficient part of the general term, remembering to include the \( (-1)^r \) factor if applicable. This will give you the exact numerical coefficient.

🎯 Exam Tip: Carefully handle negative exponents and the \( (-1)^r \) factor when simplifying the general term. A small error in calculations can lead to an incorrect sign or magnitude.

Question 8. If in the expansion of \( (1 + x)^{18} \), coefficients of \( (2r + 4)^{th} \) and \( (r - 2)^{th} \) terms are equal then value of r is:
(A) 5
(B) 6
(C) 7
(D) 8
Answer: (C) 7
In simple words: When two coefficients in a binomial expansion are equal, it means that either their 'r' values are the same, or they add up to the total power 'n'. By setting the 'r' values of the given terms equal to each other or summing them to 18, we can solve for 'r'. This property helps us find the specific position.

🎯 Exam Tip: Remember the property that if \( {}^nC_x = {}^nC_y \), then either \( x = y \) or \( x + y = n \). Apply this property carefully to find the possible values of 'r'.

Question 11. Find the value of term independent of x in the expansion of \( \left(2x - \frac{1}{x}\right)^{10} \)
Answer: The general term for the expansion of \( (a+b)^n \) is \( T_{r+1} = {}^nC_r a^{n-r} b^r \).
For \( \left(2x - \frac{1}{x}\right)^{10} \), we have \( a = 2x \), \( b = -\frac{1}{x} \), and \( n = 10 \).
So, the \( (r+1)^{th} \) term is:
\( T_{r+1} = {}^{10}C_r (2x)^{10-r} \left(-\frac{1}{x}\right)^r \)
\( T_{r+1} = {}^{10}C_r 2^{10-r} x^{10-r} (-1)^r x^{-r} \)
\( T_{r+1} = {}^{10}C_r (-1)^r 2^{10-r} x^{10-r-r} \)
\( T_{r+1} = {}^{10}C_r (-1)^r 2^{10-r} x^{10-2r} \)
For the term independent of x, the power of x must be 0.
So, \( 10 - 2r = 0 \)
\( 2r = 10 \)
\( r = 5 \)
Now, substitute \( r=5 \) back into the expression for \( T_{r+1} \) (excluding the x term):
\( T_{5+1} = {}^{10}C_5 (-1)^5 2^{10-5} \)
\( T_6 = \frac{10!}{5!(10-5)!} (-1) 2^5 \)
\( T_6 = \frac{10!}{5!5!} (-1) 32 \)
\( T_6 = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5 \times 4 \times 3 \times 2 \times 1 \times 5!} (-1) 32 \)
\( T_6 = (2 \times 3 \times 2 \times 7 \times 3) (-1) 32 \)
\( T_6 = 252 (-1) 32 \)
\( T_6 = -8064 \)
Thus, the value of the term independent of x is \( -8064 \).
In simple words: To find the part of the expansion that doesn't have 'x', we first write down the general formula for any term. Then, we figure out what 'r' needs to be so that the 'x' part completely disappears. After finding 'r', we put it back into the formula to get the final numerical answer. The term independent of x is simply a constant number.

🎯 Exam Tip: Always set the exponent of 'x' in the general term \( T_{r+1} \) to zero to find the term independent of 'x'. Be careful with signs, especially when there's a negative term in the binomial.

Question 13. If the expansion of \( (1 + x)^n \), \( C_0 + C_1 + C_2 + C_3 + ... C_n \) are coefficients different terms then find the value \( C_0 + C_2 + C_4... \)
Answer: We know the binomial expansion for \( (1+x)^n \) is:
\( (1+x)^n = {}^nC_0 + {}^nC_1 x + {}^nC_2 x^2 + {}^nC_3 x^3 + ... + {}^nC_n x^n \)
Let's call the coefficients \( C_0, C_1, C_2, ... C_n \).
So, \( (1+x)^n = C_0 + C_1 x + C_2 x^2 + C_3 x^3 + ... + C_n x^n \) (Equation i)
If we put \( x=1 \) in Equation (i):
\( (1+1)^n = C_0 + C_1 (1) + C_2 (1)^2 + C_3 (1)^3 + ... + C_n (1)^n \)
\( 2^n = C_0 + C_1 + C_2 + C_3 + ... + C_n \) (Equation A)
If we put \( x=-1 \) in Equation (i):
\( (1+(-1))^n = C_0 + C_1 (-1) + C_2 (-1)^2 + C_3 (-1)^3 + ... + C_n (-1)^n \)
\( 0^n = C_0 - C_1 + C_2 - C_3 + ... + (-1)^n C_n \)
\( 0 = C_0 - C_1 + C_2 - C_3 + ... \) (Equation B)
Now, let's add Equation (A) and Equation (B):
\( (C_0 + C_1 + C_2 + C_3 + ...) + (C_0 - C_1 + C_2 - C_3 + ...) = 2^n + 0 \)
\( 2C_0 + 2C_2 + 2C_4 + ... = 2^n \)
\( 2(C_0 + C_2 + C_4 + ...) = 2^n \)
Divide both sides by 2:
\( C_0 + C_2 + C_4 + ... = \frac{2^n}{2} \)
\( C_0 + C_2 + C_4 + ... = 2^{n-1} \)
Therefore, the sum of the even-indexed coefficients is \( 2^{n-1} \). This result is a very useful property of binomial coefficients.
In simple words: We are asked to find the sum of coefficients for terms with even powers of 'x' in a binomial expansion. We use the binomial theorem and substitute \( x=1 \) and \( x=-1 \) into the expansion. By adding these two resulting equations, all the odd-indexed coefficients cancel out, leaving us with double the sum of the even-indexed coefficients, which simplifies to \( 2^{n-1} \).

🎯 Exam Tip: Remember the property that the sum of even-indexed coefficients is equal to the sum of odd-indexed coefficients, and both are \( 2^{n-1} \). This is a common shortcut for binomial coefficient problems.

Question 14. Find the value of \( {}^{30}C_1 + {}^{30}C_2 + {}^{30}C_3 +... + {}^{30}C_{30} \).
Answer: We know a general property of binomial coefficients that the sum of all coefficients in the expansion of \( (1+x)^n \) is \( 2^n \).
This means: \( {}^{n}C_0 + {}^{n}C_1 + {}^{n}C_2 + ... + {}^{n}C_n = 2^n \).
In this question, \( n = 30 \). So, the sum of all coefficients would be:
\( {}^{30}C_0 + {}^{30}C_1 + {}^{30}C_2 + ... + {}^{30}C_{30} = 2^{30} \)
We are asked to find the value of \( {}^{30}C_1 + {}^{30}C_2 + {}^{30}C_3 +... + {}^{30}C_{30} \).
Notice that the term \( {}^{30}C_0 \) is missing from the required sum. We know that \( {}^{n}C_0 = 1 \) for any n.
So, \( {}^{30}C_0 = 1 \).
We can rewrite the equation as:
\( 1 + ({}^{30}C_1 + {}^{30}C_2 + {}^{30}C_3 +... + {}^{30}C_{30}) = 2^{30} \)
Now, to find the required sum, we subtract 1 from \( 2^{30} \):
\( {}^{30}C_1 + {}^{30}C_2 + {}^{30}C_3 +... + {}^{30}C_{30} = 2^{30} - 1 \)
This property is very useful for quick calculations involving sums of binomial coefficients.
In simple words: The total sum of all possible binomial coefficients for a power 'n' is \( 2^n \). This question asks for the sum of all coefficients except the first one, \( {}^{30}C_0 \). Since \( {}^{30}C_0 \) is always 1, we just subtract 1 from the total sum \( 2^{30} \) to get the answer.

🎯 Exam Tip: Always remember that \( {}^{n}C_0 = 1 \) and \( {}^{n}C_n = 1 \). The sum of all binomial coefficients for power n is \( 2^n \). If the sum starts from \( {}^{n}C_1 \) instead of \( {}^{n}C_0 \), subtract 1 from \( 2^n \).

Question 15. Find the middle term in the expansion of \( \left(\frac{a}{x} + \frac{x}{a}\right)^{10} \).
Answer: In the binomial expansion of \( (A+B)^n \), if 'n' is an even number, there is only one middle term. Its position is given by \( \left(\frac{n}{2} + 1\right)^{th} \).
Here, \( n = 10 \), which is an even number.
So, the position of the middle term is \( \left(\frac{10}{2} + 1\right)^{th} = (5+1)^{th} = 6^{th} \) term.
We need to find the \( 6^{th} \) term, which means \( r = 5 \) (since \( T_{r+1} \)).
Using the general term formula \( T_{r+1} = {}^nC_r A^{n-r} B^r \):
\( T_6 = {}^{10}C_5 \left(\frac{a}{x}\right)^{10-5} \left(\frac{x}{a}\right)^5 \)
\( T_6 = {}^{10}C_5 \left(\frac{a}{x}\right)^5 \left(\frac{x}{a}\right)^5 \)
\( T_6 = {}^{10}C_5 \left(\frac{a^5}{x^5}\right) \left(\frac{x^5}{a^5}\right) \)
\( T_6 = {}^{10}C_5 \times 1 \)
Now, calculate \( {}^{10}C_5 \):
\( {}^{10}C_5 = \frac{10!}{5!(10-5)!} = \frac{10!}{5!5!} \)
\( {}^{10}C_5 = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5 \times 4 \times 3 \times 2 \times 1 \times 5!} \)
\( {}^{10}C_5 = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \)
\( {}^{10}C_5 = 2 \times 3 \times 2 \times 7 \times 3 \)
\( {}^{10}C_5 = 252 \)
So, the middle term is 252. Interestingly, this expansion only yields a constant middle term.
In simple words: For a binomial expression raised to an even power, the middle term is found by taking half the power and adding one to find its position. Then, we use the binomial formula to calculate that specific term. In this case, the 6th term is the middle term, and after calculations, it turns out to be just the number 252.

🎯 Exam Tip: When \( n \) is even, the middle term is \( T_{\frac{n}{2}+1} \). If the terms are reciprocals like \( \frac{a}{x} \) and \( \frac{x}{a} \), their powers will often cancel out in the middle term, leaving a constant.

Question 16. In the product of expansion of \( (1 + 2x)^6 (1 - x)^7 \), find the coefficient of \( x^5 \).
Answer: We need to find the coefficient of \( x^5 \) in the product of \( (1 + 2x)^6 \) and \( (1 - x)^7 \).
First, let's write out the expansions of both terms using the binomial theorem \( (a+b)^n = \sum_{r=0}^n {}^nC_r a^{n-r} b^r \).
For \( (1 + 2x)^6 \):
\( (1 + 2x)^6 = {}^6C_0 (1)^6 (2x)^0 + {}^6C_1 (1)^5 (2x)^1 + {}^6C_2 (1)^4 (2x)^2 + {}^6C_3 (1)^3 (2x)^3 + {}^6C_4 (1)^2 (2x)^4 + {}^6C_5 (1)^1 (2x)^5 + {}^6C_6 (1)^0 (2x)^6 \)
\( = 1 + 6(2x) + 15(4x^2) + 20(8x^3) + 15(16x^4) + 6(32x^5) + 1(64x^6) \)
\( = 1 + 12x + 60x^2 + 160x^3 + 240x^4 + 192x^5 + 64x^6 \) (Equation i)
For \( (1 - x)^7 \):
\( (1 - x)^7 = {}^7C_0 (1)^7 (-x)^0 + {}^7C_1 (1)^6 (-x)^1 + {}^7C_2 (1)^5 (-x)^2 + {}^7C_3 (1)^4 (-x)^3 + {}^7C_4 (1)^3 (-x)^4 + {}^7C_5 (1)^2 (-x)^5 + {}^7C_6 (1)^1 (-x)^6 + {}^7C_7 (1)^0 (-x)^7 \)
\( = 1 - 7x + 21x^2 - 35x^3 + 35x^4 - 21x^5 + 7x^6 - x^7 \) (Equation ii)
Now, we need to find the coefficient of \( x^5 \) when we multiply (Equation i) and (Equation ii). We look for pairs of terms whose powers of x add up to 5:
\( (1)( -21x^5 ) \Rightarrow -21x^5 \)
\( (12x)( 35x^4 ) \Rightarrow 420x^5 \)
\( (60x^2)( -35x^3 ) \Rightarrow -2100x^5 \)
\( (160x^3)( 21x^2 ) \Rightarrow 3360x^5 \)
\( (240x^4)( -7x ) \Rightarrow -1680x^5 \)
\( (192x^5)( 1 ) \Rightarrow 192x^5 \)
Adding all these coefficients together:
Coefficient of \( x^5 = -21 + 420 - 2100 + 3360 - 1680 + 192 \)
\( = (-21 - 2100 - 1680) + (420 + 3360 + 192) \)
\( = -3801 + 3972 \)
\( = 171 \)
The coefficient of \( x^5 \) in the product of the expansions is 171.
In simple words: To find the \( x^5 \) part when multiplying two expansions, we first expand each part separately. Then, we find all the combinations of terms from the two expansions whose powers of 'x' add up to 5. We multiply the coefficients for each of these pairs and then add up all those products to get the final coefficient for \( x^5 \).

🎯 Exam Tip: When finding the coefficient of a specific power in a product of expansions, list out the relevant terms from each expansion, then systematically combine them to ensure no pair is missed. Be very careful with signs and calculations.

Question 17. If in the expansion of \( (1 + x)^{2n} \) coefficient of 2nd, 3rd and 4th terms are in A.P. then prove that \( 2n^2 – 9n + 7 = 0 \).
Answer: If three terms a, b, c are in Arithmetic Progression (A.P.), then \( 2b = a + c \).
In the expansion of \( (1 + x)^{2n} \), the coefficients of the terms are given by \( {}^N C_r \), where \( N = 2n \).
Coefficient of the 2nd term \( (T_2) \) is \( {}^{2n}C_1 \).
Coefficient of the 3rd term \( (T_3) \) is \( {}^{2n}C_2 \).
Coefficient of the 4th term \( (T_4) \) is \( {}^{2n}C_3 \).
Since these coefficients are in A.P., we have:
\( 2 \times ({}^{2n}C_2) = {}^{2n}C_1 + {}^{2n}C_3 \)
Now, let's write out the combinations:
\( {}^N C_r = \frac{N!}{r!(N-r)!} \)
So, \( 2 \times \frac{(2n)!}{2!(2n-2)!} = \frac{(2n)!}{1!(2n-1)!} + \frac{(2n)!}{3!(2n-3)!} \)
Divide the entire equation by \( (2n)! \) (since \( (2n)! \) cannot be zero):
\( 2 \times \frac{1}{2!(2n-2)!} = \frac{1}{1!(2n-1)!} + \frac{1}{3!(2n-3)!} \)
\( \frac{1}{(2n-2)!} = \frac{1}{(2n-1)(2n-2)!} + \frac{1}{3 \times 2 \times 1 \times (2n-3)!} \)
\( \frac{1}{(2n-2)!} = \frac{1}{(2n-1)(2n-2)!} + \frac{1}{6(2n-3)!} \)
Multiply the whole equation by \( (2n-2)! \):
\( 1 = \frac{1}{2n-1} + \frac{(2n-2)!}{6(2n-3)!} \)
We know that \( (2n-2)! = (2n-2)(2n-3)! \). So:
\( 1 = \frac{1}{2n-1} + \frac{(2n-2)(2n-3)!}{6(2n-3)!} \)
\( 1 = \frac{1}{2n-1} + \frac{2n-2}{6} \)
Now, find a common denominator and solve for n:
\( 1 = \frac{6 + (2n-2)(2n-1)}{6(2n-1)} \)
\( 6(2n-1) = 6 + (4n^2 - 2n - 4n + 2) \)
\( 12n - 6 = 6 + 4n^2 - 6n + 2 \)
\( 12n - 6 = 4n^2 - 6n + 8 \)
Move all terms to one side to form a quadratic equation:
\( 0 = 4n^2 - 6n - 12n + 8 + 6 \)
\( 0 = 4n^2 - 18n + 14 \)
Divide the entire equation by 2:
\( 0 = 2n^2 - 9n + 7 \)
This proves that \( 2n^2 - 9n + 7 = 0 \). The steps involve careful simplification of factorials and algebraic manipulation.
In simple words: When terms are in an arithmetic progression, the middle term is the average of its neighbors. We apply this rule to the coefficients of the 2nd, 3rd, and 4th terms of the given binomial expansion. By writing out these coefficients using their factorial formulas and simplifying the equation, we can show that the given quadratic equation for 'n' is true.

🎯 Exam Tip: Remember the A.P. condition: \( 2b = a + c \). Be systematic and careful when expanding factorials and performing algebraic simplification to avoid errors. \( n! = n \times (n-1)! \).

Question 18. Find the expansion of \( \left(1+\frac{x}{2}-\frac{2}{x}\right)^4 \), \( x \neq 0 \) using Binomial theorem.
Answer: We need to expand \( \left(1+\frac{x}{2}-\frac{2}{x}\right)^4 \). We can treat \( \left(\frac{x}{2}-\frac{2}{x}\right) \) as a single term, say 'y', to apply the binomial theorem.
Let \( y = \frac{x}{2}-\frac{2}{x} \).
So, the expression becomes \( (1+y)^4 \).
Using the binomial theorem for \( (1+y)^4 \):
\( (1+y)^4 = {}^4C_0 (1)^4 y^0 + {}^4C_1 (1)^3 y^1 + {}^4C_2 (1)^2 y^2 + {}^4C_3 (1)^1 y^3 + {}^4C_4 (1)^0 y^4 \)
\( (1+y)^4 = 1 + 4y + 6y^2 + 4y^3 + y^4 \)
Now, substitute \( y = \left(\frac{x}{2}-\frac{2}{x}\right) \) back into the expansion:
\( 1 + 4\left(\frac{x}{2}-\frac{2}{x}\right) + 6\left(\frac{x}{2}-\frac{2}{x}\right)^2 + 4\left(\frac{x}{2}-\frac{2}{x}\right)^3 + \left(\frac{x}{2}-\frac{2}{x}\right)^4 \)

Let's expand each term:
**Term 1:** \( 1 \)

**Term 2:** \( 4\left(\frac{x}{2}-\frac{2}{x}\right) = 4 \times \frac{x}{2} - 4 \times \frac{2}{x} = 2x - \frac{8}{x} \)

**Term 3:** \( 6\left(\frac{x}{2}-\frac{2}{x}\right)^2 = 6\left[\left(\frac{x}{2}\right)^2 - 2\left(\frac{x}{2}\right)\left(\frac{2}{x}\right) + \left(\frac{2}{x}\right)^2\right] \)
\( = 6\left[\frac{x^2}{4} - 2 + \frac{4}{x^2}\right] = \frac{6x^2}{4} - 12 + \frac{24}{x^2} = \frac{3x^2}{2} - 12 + \frac{24}{x^2} \)

**Term 4:** \( 4\left(\frac{x}{2}-\frac{2}{x}\right)^3 = 4\left[\left(\frac{x}{2}\right)^3 - 3\left(\frac{x}{2}\right)^2\left(\frac{2}{x}\right) + 3\left(\frac{x}{2}\right)\left(\frac{2}{x}\right)^2 - \left(\frac{2}{x}\right)^3\right] \)
\( = 4\left[\frac{x^3}{8} - 3\frac{x^2}{4}\frac{2}{x} + 3\frac{x}{2}\frac{4}{x^2} - \frac{8}{x^3}\right] \)
\( = 4\left[\frac{x^3}{8} - \frac{3x}{2} + \frac{6}{x} - \frac{8}{x^3}\right] \)
\( = \frac{4x^3}{8} - \frac{12x}{2} + \frac{24}{x} - \frac{32}{x^3} = \frac{x^3}{2} - 6x + \frac{24}{x} - \frac{32}{x^3} \)

**Term 5:** \( \left(\frac{x}{2}-\frac{2}{x}\right)^4 \)
\( = \left[\left(\frac{x}{2}\right)^2 - 2 + \left(\frac{2}{x}\right)^2\right]^2 \) (using the expanded form of \( y^2 \))
\( = \left[\frac{x^2}{4} - 2 + \frac{4}{x^2}\right]^2 \)
\( = \left(\frac{x^2}{4}\right)^2 + (-2)^2 + \left(\frac{4}{x^2}\right)^2 + 2\left(\frac{x^2}{4}\right)(-2) + 2(-2)\left(\frac{4}{x^2}\right) + 2\left(\frac{x^2}{4}\right)\left(\frac{4}{x^2}\right) \)
\( = \frac{x^4}{16} + 4 + \frac{16}{x^4} - x^2 - \frac{16}{x^2} + 2 \)
\( = \frac{x^4}{16} - x^2 + 6 - \frac{16}{x^2} + \frac{16}{x^4} \)

Now, sum all the expanded terms:
\( \left(1+\frac{x}{2}-\frac{2}{x}\right)^4 = 1 + \left(2x - \frac{8}{x}\right) + \left(\frac{3x^2}{2} - 12 + \frac{24}{x^2}\right) + \left(\frac{x^3}{2} - 6x + \frac{24}{x} - \frac{32}{x^3}\right) + \left(\frac{x^4}{16} - x^2 + 6 - \frac{16}{x^2} + \frac{16}{x^4}\right) \)

Group terms by powers of x:
**Constant terms:** \( 1 - 12 + 6 = -5 \)
**\( x^1 \) terms:** \( 2x - 6x = -4x \)
**\( x^2 \) terms:** \( \frac{3x^2}{2} - x^2 = \frac{3x^2 - 2x^2}{2} = \frac{x^2}{2} \)
**\( x^3 \) terms:** \( \frac{x^3}{2} \)
**\( x^4 \) terms:** \( \frac{x^4}{16} \)
**\( x^{-1} \) terms (i.e., \( \frac{1}{x} \)):** \( -\frac{8}{x} + \frac{24}{x} = \frac{16}{x} \)
**\( x^{-2} \) terms (i.e., \( \frac{1}{x^2} \)):** \( \frac{24}{x^2} - \frac{16}{x^2} = \frac{8}{x^2} \)
**\( x^{-3} \) terms (i.e., \( \frac{1}{x^3} \)):** \( -\frac{32}{x^3} \)
**\( x^{-4} \) terms (i.e., \( \frac{1}{x^4} \)):** \( \frac{16}{x^4} \)

Combining all terms in ascending order of power of x:
\( \frac{x^4}{16} + \frac{x^3}{2} + \frac{x^2}{2} - 4x - 5 + \frac{16}{x} + \frac{8}{x^2} - \frac{32}{x^3} + \frac{16}{x^4} \)
This expansion requires careful application of the binomial theorem multiple times and thorough algebraic simplification.
In simple words: To expand an expression with three terms raised to a power, we first group two terms together and treat them as a single variable. Then, we apply the binomial theorem. After expanding, we replace the grouped variable with its original terms and expand again, carefully combining all similar terms (terms with the same power of x) to get the final, simplified expansion.

🎯 Exam Tip: When expanding trinomials, use substitution to simplify the problem into a binomial expansion. Be methodical in expanding each substituted term and meticulous in combining like terms, paying close attention to signs and denominators.