If x = frac{1}{5} + frac{1 times 3}{5 times 1 | vedclass

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(B) The given series is $x = \frac{1}{5} + \frac{1 \cdot 3}{5 \cdot 10} + \frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15} + \ldots$ We know the binomial

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$2$

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(B) The given series is $x = \frac{1}{5} + \frac{1 \cdot 3}{5 \cdot 10} + \frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15} + \ldots$ We know the binomial expansion $(1-y)^{-n} = 1 + ny + \frac{n(n+1)}{2!}y^2 + \frac{n(n+1)(n+2)}{3!}y^3 + \ldots$ Let $1+x = 1 + \frac{1}{5} + \frac{1 \cdot 3}{5 \cdot 10} + \frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15} + \ldots$ This matches the form $(1-y)^{-n}$ where $ny = \frac{1}{5}$ and $\frac{n(n+1)}{2}y^2 = \frac{3}{50}$. Solving for $n$ and $y$,we find $n = \frac{1}{2}$ and $y = \frac{2}{5}$. Thus,$1+x = (1 - \frac{2}{5})^{-1/2} = (\frac{3}{5})^{-1/2} = \sqrt{\frac{5}{3}}$. So,$x = \sqrt{\frac{5}{3}} - 1$. Now,calculate $3x^2 + 6x = 3(x^2 + 2x) = 3((x+1)^2 - 1) = 3(x+1)^2 - 3$. Substituting $x+1 = \sqrt{\frac{5}{3}}$,we get $3(\frac{5}{3}) - 3 = 5 - 3 = 2$.

If $x = \frac{1}{5} + \frac{1 \times 3}{5 \times 10} + \frac{1 \times 3 \times 5}{5 \times 10 \times 15} + \ldots$,then $3x^2 + 6x =$

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