If the Coefficient of x³⁰ in the expansion of (1 + 1/x)⁶ (1 + x²)⁷ (1 - x³)⁸; x ≠ 0 is α, then |α| e

Q: Why did the target power change from $x^{30}$ to $x^{36}$?

Because we simplified the first term to $\frac{(1+x)^6}{x^6}$. Since we are dividing by $x^6$, we need the numerator to provide an $x^{36}$ term so that $\frac{x^{36}}{x^6} = x^{30}$. Always watch the denominator!

Q: How do we know which values of $r_3$ to test?

We use bounding logic. The maximum value of $r_1 + 2r_2$ is $6 + 2(7) = 20$. Since $r_1 + 2r_2 + 3r_3 = 36$, we have $3r_3 = 36 - (r_1 + 2r_2)$. The minimum value of $3r_3$ is $36 - 20 = 16$, so $r_3 \ge 5.33$. Since $r_3$ is an integer and $r_3 \le 8$, we only check $6, 7, 8$.

Q: Why does the sign change in Case 2?

The third bracket is $(1 - x^3)^8$. The general term involves $(-x^3)^{r_3}$, which is $(-1)^{r_3} \cdot x^{3r_3}$. When $r_3$ is odd (like 7), the term is negative. When $r_3$ is even (like 6 or 8), the term is positive. Always keep track of that $(-1)^{r_3}$ factor!

Question

If the Coefficient of $x^{30}$ in the expansion of $(1 + \frac{1}{x})^6 (1 + x^2)^7 (1 - x^3)^8; x 
eq 0$ is $\alpha$, then $|\alpha|$ equals

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the coefficient of $x^{30}$ in the expansion of the expression:
$$(1 + \frac{1}{x})^6 (1 + x^2)^7 (1 - x^3)^8$$

The term $(1 + \frac{1}{x})^6$ introduces a denominator that complicates the expansion. We simplify this by rewriting it as:
$$\frac{(1+x)^6}{x^6}$$

Substituting this back into the original expression, we obtain:
$$\frac{(1+x)^6 (1+x^2)^7 (1-x^3)^8}{x^6}$$

To find the coefficient of $x^{30}$ in the original expression, we must now find the coefficient of $x^{36}$ in the numerator, since:
$$\frac{x^{36}}{x^6} = x^{30}$$

### The Detective Work: Setting the Constraints

We define the general term for each binomial factor:
1. From $(1+x)^6$: $\binom{6}{r_1} x^{r_1}$ where $0 \le r_1 \le 6$
2. From $(1+x^2)^7$: $\binom{7}{r_2} (x^2)^{r_2}$ where $0 \le r_2 \le 7$
3. From $(1-x^3)^8$: $\binom{8}{r_3} (-x^3)^{r_3}$ where $0 \le r_3 \le 8$

Multiplying these terms, the power of $x$ is given by $r_1 + 2r_2 + 3r_3$. We set this equal to $36$:
$$r_1 + 2r_2 + 3r_3 = 36$$

Given the constraints $r_1 \le 6$ and $r_2 \le 7$, the maximum value of $r_1 + 2r_2$ is $6 + 2(7) = 20$. Consequently, $3r_3$ must be at least $36 - 20 = 16$. This restricts $r_3$ to the values $\{6, 7, 8\}$.

### The Final Calculation

We evaluate the cases for $r_3$ systematically:

**Case 1: $r_3 = 8$**
The equation becomes $r_1 + 2r_2 = 36 - 24 = 12$.
Possible $(r_1, r_2)$ pairs are $(0, 6), (2, 5), (4, 4), (6, 3)$.
The sum of coefficients is $\binom{8}{8} 	imes [\binom{6}{0}\binom{7}{6} + \binom{6}{2}\binom{7}{5} + \binom{6}{4}\binom{7}{4} + \binom{6}{6}\binom{7}{3}] = 1 	imes [7 + 315 + 525 + 35] = 882$.

**Case 2: $r_3 = 7$**
The equation becomes $r_1 + 2r_2 = 36 - 21 = 15$.
Possible $(r_1, r_2)$ pairs are $(1, 7), (3, 6), (5, 5)$.
Since $r_3$ is odd, the term $(-1)^{r_3}$ is $-1$.
The sum of coefficients is $-\binom{8}{7} 	imes [\binom{6}{1}\binom{7}{7} + \binom{6}{3}\binom{7}{6} + \binom{6}{5}\binom{7}{5}] = -8 	imes [6 + 140 + 126] = -8 	imes 272 = -2176$.

**Case 3: $r_3 = 6$**
The equation becomes $r_1 + 2r_2 = 36 - 18 = 18$.
Possible $(r_1, r_2)$ pairs are $(4, 7), (6, 6)$.
The sum of coefficients is $\binom{8}{6} 	imes [\binom{6}{4}\binom{7}{7} + \binom{6}{6}\binom{7}{6}] = 28 	imes [15 + 7] = 28 	imes 22 = 616$.

Summing these results: $882 - 2176 + 616 = -678$.
The absolute value of the coefficient is **$678$**.

Animated Solution for Mathematics - Binomial Theorem: If the Coefficient of $x^{30}$ in the expansion of $(1 + \frac{1}{x})^{6} (1 + x^{2})^{7} (1 - x^{3})^{8}; x \neq = 0$ is $α$ , then $∣ α ∣$ equals

Enter Numerical Value:

Visualized Solution

Problem Analysis & Objective

Simplifying the Fractional Term

Redefining the Target Power

Setting up the General Term

The Power Equation

Constraint Analysis

Case 1: $r_{3} = 8$

Calculating Coefficients for Case 1

Case 2: $r_{3} = 7$

Calculating Coefficients for Case 2

Case 3: $r_{3} = 6$

Calculating Coefficients for Case 3

Final Summation & Result

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The Detective Work

The Final Calculation

Similar Questions

Let $α > 0$ , be the smallest number such that the expansion of $(x^{2/3} + \frac{2}{x ^{3}})^{30}$ has a term $β x^{- α}, β \in N$ . Then $α$ is equal to ______.

Let $α$ be the constant term in the binomial expansion of $(x - \frac{6}{x ^{\frac{3}{2}}})^{n}, n \leq 15$ . If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of $x^{- n}$ is $λ α$ , then $λ$ is equal to ————————.

The Coefficient of $x^{- 6}$ in the expansion of $(\frac{4 x}{5} + \frac{5}{2 x ^{2}})^{9}$ is ______.

If the constant term in the expansion of $(\frac{5 3}{x} + \frac{2 x}{3 5})^{12}, x \neq = 0$ , is $α \times 2^{8} im es 53$ , then $25 α$ is equal to :

If the co-efficient of $x^{7}$ and $x^{8}$ in the expansion of $(2 + \frac{x}{3})^{n}$ are equal, then the value of $n$ is equal to ___

The coefficient of $x^{10}$ in the expansion of $(1 + x)^{2} (1 + x^{2})^{3} (1 + x^{3})^{4}$ is equal to :-

If the constant term in the expansion of $(1 + 2 x - 3 x^{3}) (\frac{3}{2} x^{2} - \frac{1}{3 x})^{9}$ is $p$ , then $108 p$ is equal to

The coefficient of $x^{5}$ in the expansion of $(2 x^{3} - \frac{1}{3 x ^{2}})^{5}$ is

If the constant term in the binomial expansion of $(\frac{x ^{5/2}}{2} - \frac{4}{x ^{l}})^{9}$ is $- 84$ and the Coefficient of $x^{- 3 l}$ is $2^{α} β$ , where $β < 0$ is an odd number, Then $∣ α l - β ∣$ is equal to ______.

The coefficient of the middle term in the binomial expansion in powers of $x$ of $(1 + α x)^{4}$ and of $(1 - α x)^{6}$ is the same if $α$ equals

Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the Coefficient of x30 in the expansion of (1+x1​)6(1+x2)7(1−x3)8;x=0 is α, then ∣α∣ equals