Expression Simplification
First, simplify the expression inside the parenthesis: $ \left(\frac{x+1}{x^{\frac{2}{3}} - x^{\frac{1}{3}} + 1} - \frac{x-1}{x - x^{\frac{1}{2}}}\right)^{15} $ The first term, $ \frac{x+1}{x^{\frac{2}{3}} - x^{\frac{1}{3}} + 1} $, simplifies using the sum of cubes identity $a^3+b^3 = (a+b)(a^2-ab+b^2)$. With $a = x^{\frac{1}{3}}$ and $b=1$, this term becomes $a+b = x^{\frac{1}{3}} + 1$. The second term, $ \frac{x-1}{x - x^{\frac{1}{2}}} $, simplifies by factoring: $ \frac{(x^{\frac{1}{2}}-1)(x^{\frac{1}{2}}+1)}{x^{\frac{1}{2}}(x^{\frac{1}{2}}-1)} = \frac{x^{\frac{1}{2}}+1}{x^{\frac{1}{2}}} = 1 + x^{-\frac{1}{2}} $. Combining these, the expression inside the parenthesis is $ (x^{\frac{1}{3}} + 1) - (1 + x^{-\frac{1}{2}}) = x^{\frac{1}{3}} - x^{-\frac{1}{2}} $.
Binomial Expansion Setup
We need the term independent of $x$ in the expansion of $ \left( x^{\frac{1}{3}} - x^{-\frac{1}{2}} \right)^{15} $. The general term $T_{r+1}$ in the binomial expansion of $ (a+b)^n $ is $ T_{r+1} = \binom{n}{r} a^{n-r} b^r $. For this problem, $n=15$, $a = x^{\frac{1}{3}}$, and $b = -x^{-\frac{1}{2}}$. $ T_{r+1} = \binom{15}{r} \left( x^{\frac{1}{3}} \right)^{15-r} \left( -x^{-\frac{1}{2}} \right)^r $ $ T_{r+1} = \binom{15}{r} (-1)^r x^{\frac{15-r}{3}} x^{-\frac{r}{2}} $ Combine the exponents of $x$: $ T_{r+1} = \binom{15}{r} (-1)^r x^{\frac{15-r}{3} - \frac{r}{2}} $
Solving for the Term Index
The term is independent of $x$ when the exponent of $x$ is zero. $ \frac{15-r}{3} - \frac{r}{2} = 0 $ Multiplying by 6 yields: $ 2(15-r) - 3r = 0 $ $ 30 - 2r - 3r = 0 $ $ 30 - 5r = 0 $ $ 5r = 30 \implies r = 6 $
Calculating the Constant Term
The term independent of $x$ corresponds to $r=6$, which is the $T_7$ term. $ T_7 = \binom{15}{6} (-1)^6 x^{\frac{15-6}{3} - \frac{6}{2}} $ $ T_7 = \binom{15}{6} \times 1 \times x^0 = \binom{15}{6} $ Calculate the binomial coefficient: $ \binom{15}{6} = \frac{15!}{6!(15-6)!} = \frac{15!}{6!9!} $ $ \binom{15}{6} = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10}{6 \times 5 \times 4 \times 3 \times 2 \times 1} $ Simplify the calculation: $ \binom{15}{6} = \frac{15}{5 \times 3} \times \frac{14}{2} \times 13 \times \frac{12}{6 \times 4} \times 11 \times 10 \text{ (Incorrect simplification method)} $ $ \binom{15}{6} = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10}{720} $ Cancel terms: $ \frac{12}{6 \times 2} = 1 $ and $ \frac{15}{5 \times 3} = 1 $. $ \binom{15}{6} = \frac{1 \times 14 \times 13 \times 1 \times 11 \times 10}{4 \times 1} = \frac{14 \times 13 \times 11 \times 10}{4} $ Simplify further: $ \frac{10}{4} = \frac{5}{2} $. $ \binom{15}{6} = \frac{14 \times 13 \times 11 \times 5}{2} $ Simplify $ \frac{14}{2} = 7 $. $ \binom{15}{6} = 7 \times 13 \times 11 \times 5 = 91 \times 55 = 5005 $ The term independent of $x$ is $5005$.
Similar Questions
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Important Questions from Algebra
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
- The number of $3 \times 2$ matrices A, which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^T A$ is 5, is ______
- The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is ______
- Let $z = (1 + i)(1 + 2i)(1 + 3i) \dots (1 + ni)$, where $i = \sqrt{-1}$. If $|z|^2 = 44200$, then $n$ is equal to _______
- Let $S = \{(m, n) : m, n \in \{1, 2, 3, \dots, 50\}\}$. If the number of elements $(m, n)$ in S such that $6^m + 9^n$ is a multiple of 5 is p and the number of elements $(m, n)$ in S such that $m + n$ is a square of a prime number is q, then $p + q$ is equal to ____________.
