If coefficient of x in expansion of (x²+k/x)⁵ is 270,find k. In binomial theorem

We are asked to find the value of \( k \) if the coefficient of \( x \) in the expansion of \( \left( x^2 + \frac{k}{x} \right)^5 \) is 270.

### Step 1: Apply the Binomial Theorem
The binomial expansion of \( (a + b)^n \) is given by:

\[
(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r
\]

In this case:
- \( a = x^2 \)
- \( b = \frac{k}{x} \)
- \( n = 5 \)

So, the general term in the expansion is:

\[
T_r = \binom{5}{r} (x^2)^{5-r} \left( \frac{k}{x} \right)^r
\]

### Step 2: Simplify the General Term
Now, simplify \( T_r \):

\[
T_r = \binom{5}{r} x^{2(5-r)} \cdot \frac{k^r}{x^r} = \binom{5}{r} k^r x^{10 - 3r}
\]

### Step 3: Find the Coefficient of \( x^1 \)
We need the term where the power of \( x \) is 1, i.e., \( 10 - 3r = 1 \). Solving for \( r \):

\[
10 - 3r = 1
\]

\[
3r = 9
\]

\[
r = 3
\]

### Step 4: Substitute \( r = 3 \) into the General Term
Now substitute \( r = 3 \) into the expression for \( T_r \):

\[
T_3 = \binom{5}{3} k^3 x^{10 - 3(3)} = \binom{5}{3} k^3 x^1
\]

We know \( \binom{5}{3} = 10 \), so:

\[
T_3 = 10 k^3 x
\]

The coefficient of \( x \) is given as 270, so:

\[
10 k^3 = 270
\]

### Step 5: Solve for \( k \)
Solving for \( k^3 \):

\[
k^3 = \frac{270}{10} = 27
\]

Taking the cube root of both sides:

\[
k = \sqrt[3]{27} = 3
\]

Thus, the value of \( k \) is \( \boxed{3} \).

---

Do you want more details or have any questions? Here are some related questions:

1. How does the binomial theorem help in expanding polynomials?
2. What are the conditions for applying the binomial theorem?
3. Can the binomial theorem be used for negative or fractional exponents?
4. What is the significance of the general term in a binomial expansion?
5. How do we identify specific powers of \( x \) in a binomial expansion?

**Tip:** In binomial expansions, finding the power of a specific term involves equating the exponent of the variable to the desired value and solving for the corresponding term index.

If the coefficient of x in the expansion of (x²+k/x)⁵ is 270, find k. In binomial theorem

Explore how to find the value of k when the coefficient of x in the expansion of (x²+k/x)⁵ is 270, using the binomial theorem. This step-by-step solution covers key algebraic concepts suitable for students in Grades 11-12.

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