An algebra problem by Sourasish Karmakar

$(1-x)^{-n}=1+nx+\dfrac{n(n+1)}{2!}x^2+\cdots+\dfrac{n(n+1)(n+2)\cdots (n+r-1)}{r!}x^{r}+\cdots\infty\quad\text{terms}$

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$1+\dfrac34+\dfrac{3\cdot5}{4\cdot8}+\dfrac{3\cdot5\cdot7}{4\cdot8\cdot12}+\cdots$

$\Rightarrow T_{r+1} = \dfrac{3\cdot5\cdot7\cdots(2r+1)}{4^{r}\times{r!}}$

$\Rightarrow T_{r+1} = \dfrac{3\cdot5\cdot7\cdots(2r+1)}{2^{r}\times{r!}}\times{\dfrac1{2^{r}}}$

$\Rightarrow T_{r+1} = \dfrac{\dfrac32\cdot\dfrac52\cdot\dfrac72\cdots\dfrac{(2r+1)}2}{r!}\times{\left(\dfrac12\right)^{r}}$

$\Rightarrow T_{r+1} = \dfrac{\dfrac32\cdot\left(\dfrac32 + 1\right)\cdot\left(\dfrac32 + 2\right)\cdots\left(\dfrac32 + r-1\right)}{r!}\times{\left(\dfrac12\right)^r}$

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Comparing with the above written binomial expansion , we get that the expression in the question is basically the expansion of $(1-x)^{-n}$ where $\boxed{x = \dfrac12 , n = \dfrac32}$

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So our answer is $\boxed{2\sqrt{2}}$