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xiaoxu200707's Question
Posted about 2 months ago

(1+x)n=1+nx+n(n1)1×2x2++n(n1)(nr+1)1×2x×rx+(x<1,nR))(1+x)12=\begin{array}{l}\left.(1+x)^{n}=1+n x+\frac{n(n-1)}{1 \times 2} x^{2}+\ldots+\frac{n(n-1) \cdot(n-r+1)}{1 \times 2 x \ldots \times r} x+\ldots(|x|<1, n \in R)\right) \\ (1+x)^{\frac{1}{2}}=\end{array}
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Answer from Sia
Posted about 2 months ago
Solution by Steps
step 1
The binomial theorem formula is given by: (1+x)n=1+nx+n(n1)1×2x2++n(n1)(nr+1)1×2××rxr+(x<1,nR) (1+x)^n = 1 + nx + \frac{n(n-1)}{1 \times 2} x^2 + \ldots + \frac{n(n-1) \cdot (n-r+1)}{1 \times 2 \times \ldots \times r} x^r + \ldots \quad (|x|<1, n \in \mathbb{R})
step 2
For the specific case where n=12 n = \frac{1}{2} , the binomial expansion becomes: (1+x)12=1+12x+12(121)1×2x2+ (1+x)^{\frac{1}{2}} = 1 + \frac{1}{2}x + \frac{\frac{1}{2}(\frac{1}{2}-1)}{1 \times 2} x^2 + \ldots
step 3
Simplifying the second term: 12(121)1×2=12122=18 \frac{\frac{1}{2}(\frac{1}{2}-1)}{1 \times 2} = \frac{\frac{1}{2} \cdot -\frac{1}{2}}{2} = -\frac{1}{8} Thus, the expansion up to the second term is: (1+x)12=1+12x18x2+ (1+x)^{\frac{1}{2}} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \ldots
The binomial expansion for (1+x)12(1+x)^{\frac{1}{2}} up to the second term is 1+12x18x21 + \frac{1}{2}x - \frac{1}{8}x^2.
Key Concept
Binomial Theorem
The binomial theorem provides a way to expand expressions of the form (1+x)n(1+x)^n for any real number nn. In this case, we used it to expand (1+x)12(1+x)^{\frac{1}{2}}.

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