$\begin{array}{l}\left.(1+x)^{n}=1+n x+\frac{n(n-1)}{1 imes 2} x^{2}+\ldots+\frac{n(n-1) \cdot(n-r+1)}{1 imes 2 x \ldots imes r} x+\ldots(|x|

Q: $\begin{array}{l}\left.(1+x)^{n}=1+n x+\frac{n(n-1)}{1 \times 2} x^{2}+\ldots+\frac{n(n-1) \cdot(n-r+1)}{1 \times 2 x \ldots \times r} x+\ldots(|x|<1, n \in R)\right) \\ (1+x)^{\frac{1}{2}}=\end{array}$

∻Solution by Steps∻ ‖ step 1 ⋮ The binomial theorem formula is given by: \[ (1+x)^n = 1 + nx + \frac{n(n-1)}{1 \times 2} x^2 + \ldots + \frac{n(n-1) \cdot (n-r+1)}{1 \times 2 \times \ldots \times r} x^r + \ldots \quad (|x|<1, n \in \mathbb{R}) \] ‖ ‖ step 2 ⋮ For the specific case where $ n = \frac{1}{2} $, the binomial expansion becomes: \[ (1+x)^{\frac{1}{2}} = 1 + \frac{1}{2}x + \frac{\frac{1}{2}(\frac{1}{2}-1)}{1 \times 2} x^2 + \ldots \] ‖ ‖ step 3 ⋮ Simplifying the second term: \[ \frac{\frac{1}{2}(\frac{1}{2}-1)}{1 \times 2} = \frac{\frac{1}{2} \cdot -\frac{1}{2}}{2} = -\frac{1}{8} \] Thus, the expansion up to the second term is: \[ (1+x)^{\frac{1}{2}} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \ldots \] ‖ ∻Answer∻ ⚹ The binomial expansion for $(1+x)^{\frac{1}{2}}$ up to the second term is $1 + \frac{1}{2}x - \frac{1}{8}x^2$. ⚹ ∻Key Concept∻ ⚹Binomial Theorem⚹ ∻Explanation∻ ⚹The binomial theorem provides a way to expand expressions of the form $(1+x)^n$ for any real number $n$. In this case, we used it to expand $(1+x)^{\frac{1}{2}}$.⚹◊What is the general formula for the binomial theorem given in the uploaded image?⍭ Generate me a similar question◊

Question

$\begin{array}{l}\left.(1+x)^{n}=1+n x+\frac{n(n-1)}{1 	imes 2} x^{2}+\ldots+\frac{n(n-1) \cdot(n-r+1)}{1 	imes 2 x \ldots 	imes r} x+\ldots(|x|<1, n \in R)ight) \ (1+x)^{\frac{1}{2}}=\end{array}$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ The binomial theorem formula is given by: $$ (1+x)^n = 1 + nx + \frac{n(n-1)}{1 \times 2} x^2 + \ldots + \frac{n(n-1) \cdot (n-r+1)}{1 \times 2 \times \ldots \times r} x^r + \ldots \quad (|x|<1, n \in \mathbb{R}) $$ ‖ ‖ step 2 ⋮ For the specific case where $ n = \frac{1}{2} $, the binomial expansion becomes: $$ (1+x)^{\frac{1}{2}} = 1 + \frac{1}{2}x + \frac{\frac{1}{2}(\frac{1}{2}-1)}{1 \times 2} x^2 + \ldots $$ ‖ ‖ step 3 ⋮ Simplifying the second term: $$ \frac{\frac{1}{2}(\frac{1}{2}-1)}{1 \times 2} = \frac{\frac{1}{2} \cdot -\frac{1}{2}}{2} = -\frac{1}{8} $$ Thus, the expansion up to the second term is: $$ (1+x)^{\frac{1}{2}} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \ldots $$ ‖ ∻Answer∻ ⚹ The binomial expansion for $(1+x)^{\frac{1}{2}}$ up to the second term is $1 + \frac{1}{2}x - \frac{1}{8}x^2$. ⚹ ∻Key Concept∻ ⚹Binomial Theorem⚹ ∻Explanation∻ ⚹The binomial theorem provides a way to expand expressions of the form $(1+x)^n$ for any real number $n$. In this case, we used it to expand $(1+x)^{\frac{1}{2}}$.⚹◊What is the general formula for the binomial theorem given in the uploaded image?⍭ Generate me a similar question◊

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