If the constant term in the expansion of (3x³ - 2x² + 5/x⁵)¹⁰ is 2ᵏ · l, where l is an odd integer,

Q: Why can't we just use the binomial theorem here?

The binomial theorem applies to expressions with only two terms, like $(a+b)^n$. Here, we have three terms: $3x^3$, $-2x^2$, and $5x^{-5}$. We must use the multinomial theorem, which generalizes the binomial approach to any number of terms.

Q: How do I know if I have found all possible solutions for $r_1, r_2, r_3$?

By expressing $r_1$ as $7r_3 - 20$, we see that $r_1$ depends entirely on $r_3$. Since $r_1 \ge 0$, we must have $7r_3 \ge 20$, meaning $r_3 \ge 3$. Also, $r_1 + r_2 + r_3 = 10$ implies $r_3$ cannot be too large. Testing $r_3=3$ and $r_3=4$ covers the valid range, and we quickly see that $r_3=4$ leads to a negative $r_2$, which is impossible.

Q: Why do we need to prime factorize 840?

The problem asks us to express the constant term in the form $2^k \cdot l$, where $l$ is an odd integer. This means we must extract all factors of $2$ from the entire expression, including the coefficient $840$ and the term $(-2)^6$. Prime factorization is the most reliable way to ensure we have captured every power of $2$.

Question

If the constant term in the expansion of $\left( 3x^3 - 2x^2 + \frac{5}{x^5} \right)^{10}$ is $2^k \cdot l$, where $l$ is an odd integer, then the value of $k$ is equal to :

Options:

(A) 6

(B) 7

(C) 8

(D) 9

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex algebraic expression, a trinomial raised to the power of ten: $(3x^3 - 2x^2 + 5x^{-5})^{10}$. While it looks daunting, we can dismantle it systematically.

Our mission is to find the **constant term**—the part of this expansion that remains untouched by the variable $x$. This is a search for a hidden symmetry within the multinomial expansion.

### The Multinomial Framework

To solve this, we invoke the multinomial theorem. The general term in the expansion of $(a+b+c)^n$ is given by:

$$T = \frac{n!}{r_1! r_2! r_3!} a^{r_1} b^{r_2} c^{r_3}$$

This is subject to the constraint $r_1 + r_2 + r_3 = n$. Here, $n=10$, $a=3x^3$, $b=-2x^2$, and $c=5x^{-5}$.

Substituting these values, the general term becomes:

$$T = \frac{10!}{r_1! r_2! r_3!} (3x^3)^{r_1} (-2x^2)^{r_2} (5x^{-5})^{r_3}$$

### The Constraint Hunt

Now, we isolate the variable $x$ by grouping the powers: $x^{3r_1 + 2r_2 - 5r_3}$. For this term to be a constant, the exponent must be zero.

This gives us our second vital equation: $3r_1 + 2r_2 - 5r_3 = 0$. We now have a system of two equations with three variables:

$$r_1 + r_2 + r_3 = 10$$
$$3r_1 + 2r_2 - 5r_3 = 0$$

By manipulating these equations, we find that $r_1 = 7r_3 - 20$. Since $r_1$ must be a non-negative integer, we immediately see that $r_3$ must be at least $3$.

Testing $r_3=3$ gives $r_1=1$, which leads to $r_2=6$. Testing $r_3=4$ leads to $r_1=8$, which forces $r_2$ to be negative—an impossibility. Thus, the only valid set is $(r_1, r_2, r_3) = (1, 6, 3)$.

### The Final Extraction

With our indices locked in, we calculate the constant term:

$$T = \frac{10!}{1! 6! 3!} (3)^1 (-2)^6 (5)^3$$

Simplifying the factorials, we find $\frac{10!}{6! 3!} = 840$. Therefore, the expression becomes:

$$T = 840 \cdot 3 \cdot 2^6 \cdot 125$$

To find $k$, we prime factorize $840 = 2^3 \cdot 3 \cdot 5 \cdot 7$. Substituting this back:

$$T = (2^3 \cdot 3 \cdot 5 \cdot 7) \cdot 3 \cdot 2^6 \cdot 5^3 = 2^9 \cdot (3^2 \cdot 5^4 \cdot 7)$$

The term in the parentheses is clearly an odd integer $l$. Comparing this to $2^k \cdot l$, we find **$k=9$**.

Animated Solution for Mathematics - Binomial Theorem: If the constant term in the expansion of $(3 x^{3} - 2 x^{2} + \frac{5}{x ^{5}})^{10}$ is $2^{k} \cdot l$ , where $l$ is an odd integer, then the value of $k$ is equal to :

Select Answer:

Visualized Solution

Understanding the Goal

Multinomial Expansion Formula

Setting up the General Term

Isolating the Power of $x$

Condition for Constant Term

Solving for $r_{1}, r_{2}, r_{3}$

Finding Valid Integer Solutions

Testing Values for $r_{3}$

Substituting $r$ Values

Simplifying the Factorials

Prime Factorization of $840$

Grouping Powers of $2$

Final Result

The Sigma Insight: Multinomial Theorem

Analyzing the Setup

The Multinomial Framework

The Constraint Hunt

The Final Extraction

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Select Answer:

Visualized Solution

Understanding the Goal

Multinomial Expansion Formula

Setting up the General Term

Isolating the Power of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Condition for Constant Term

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r1​,r2​,r3​

Finding Valid Integer Solutions

Testing Values for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r3​

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r Values

Simplifying the Factorials

Prime Factorization of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 840

Grouping Powers of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2