To solve the problem, we need to consider the expansion of \left(x^2 + \frac{k}{x}\right)^5 and find the sum of the coefficients of the x^{10}, x^4, and \frac{1}{x^2} terms, which is given as 41.

General Term in the Expansion

Using the binomial theorem, the expansion of (a+b)^n is given by:

(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \,\,\,\,\,\text{SL 1.9}

where

\binom{n}{k}=\frac{n!}{k!(n-k)!}

For our expression \left(x^2 + \frac{k}{x}\right)^5, we identify a = x^2, b = \frac{k}{x}, and n = 5. The general term T_r in the expansion is:

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

Simplifying, we have :

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

Identify Specific Terms

We need to find the coefficients of the terms where the powers of x are 10, 4, and -2.

x^{10} Term

Set the exponent of x to 10:

10 - 3r = 10 \implies 3r = 0 \implies r = 0

The coefficient of the x^{10} term is:

\binom{5}{0} k^0 = 1

x^4 Term

Set the exponent of x to 4:

10 - 3r = 4 \implies 3r = 6 \implies r = 2

The coefficient of the x^4 term is:

\binom{5}{2} k^2 = 10k^2

\frac{1}{x^2} Term

Set the exponent of x to -2:

10 - 3r = -2 \implies 3r = 12 \implies r = 4

The coefficient of the \frac{1}{x^2} term is:

\binom{5}{4} k^4 = 5k^4

Sum of the Coefficients

The sum of the coefficients of the x^{10}, x^4, and \frac{1}{x^2} terms is given as 41:

1 + 10k^2 + 5k^4 = 41

Simplifying, we get:

5k^4 + 10k^2 + 1 = 41

5k^4 + 10k^2 - 40 = 0

Divide the entire equation by 5:

k^4 + 2k^2 - 8 = 0

Let u = k^2. Then the equation becomes:

u^2 + 2u - 8 = 0

Solve the Quadratic Equation

Using the quadratic formula :

u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \,\,\,\,\,\text{SL 2.7}

where a = 1, b = 2, and c = -8:

u = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1}

u = \frac{-2 \pm \sqrt{4 + 32}}{2}

u = \frac{-2 \pm \sqrt{36}}{2}

u = \frac{-2 \pm 6}{2}

The solutions for u are:

u = \frac{4}{2} = 2 \quad \text{and} \quad u = \frac{-8}{2} = -4

Since u = k^2 and must be non-negative, we have u = 2.

Thus, k^2 = 2, giving k = \pm \sqrt{2} .

Problem Bank - Counting & Binomials

Problem Bank - Counting & Binomials