Math  /  Algebra

Question6. What is the coefficient of 1/x61 / x^{6} in the expression (x5+1/x3)10\left(x^{5}+1 / x^{3}\right)^{10}.
ANS:

Studdy Solution

STEP 1

1. We are dealing with a binomial expansion of the form (a+b)n(a + b)^n.
2. The binomial theorem can be used to expand (x5+1x3)10(x^5 + \frac{1}{x^3})^{10}.
3. We need to find the specific term in the expansion that results in x6x^{-6}.

STEP 2

1. Identify the general term in the binomial expansion.
2. Determine the power of xx in each term.
3. Set the power of xx equal to 6-6.
4. Solve for the binomial coefficient.
5. Calculate the coefficient of the desired term.

STEP 3

Identify the general term in the binomial expansion of (x5+1x3)10(x^5 + \frac{1}{x^3})^{10}.
The general term in the expansion of (a+b)n(a + b)^n is given by:
Tk=(nk)ankbk T_k = \binom{n}{k} a^{n-k} b^k
For our expression, a=x5a = x^5, b=1x3b = \frac{1}{x^3}, and n=10n = 10.
Thus, the general term TkT_k is:
Tk=(10k)(x5)10k(1x3)k T_k = \binom{10}{k} (x^5)^{10-k} \left(\frac{1}{x^3}\right)^k

STEP 4

Determine the power of xx in each term.
Tk=(10k)x5(10k)x3k T_k = \binom{10}{k} x^{5(10-k)} \cdot x^{-3k}
Combine the exponents of xx:
Tk=(10k)x505k3k=(10k)x508k T_k = \binom{10}{k} x^{50 - 5k - 3k} = \binom{10}{k} x^{50 - 8k}

STEP 5

Set the power of xx equal to 6-6 to find the appropriate term.
508k=6 50 - 8k = -6

STEP 6

Solve for kk.
508k=6 50 - 8k = -6 50+6=8k 50 + 6 = 8k 56=8k 56 = 8k k=568=7 k = \frac{56}{8} = 7

STEP 7

Calculate the coefficient of the desired term.
The coefficient is given by (107)\binom{10}{7}.
Calculate (107)\binom{10}{7}:
(107)=(103)=10×9×83×2×1=120 \binom{10}{7} = \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120
The coefficient of 1x6 \frac{1}{x^6} in the expression (x5+1/x3)10\left(x^{5}+1 / x^{3}\right)^{10} is 120 \boxed{120} .

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