If x = frac{1}{5} + frac{1 cdot 3}{5 cdot 10} | vedclass

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(B) Given that,$x = \frac{1}{5} + \frac{1 \cdot 3}{5 \cdot 10} + \frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15} + \dots$ This can be rewritten as: $x =

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(B) Given that,$x = \frac{1}{5} + \frac{1 \cdot 3}{5 \cdot 10} + \frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15} + \dots$ This can be rewritten as: $x = \frac{1}{5} + \frac{1 \cdot 3}{2!} \left(\frac{1}{5}\right)^2 + \frac{1 \cdot 3 \cdot 5}{3!} \left(\frac{1}{5}\right)^3 + \dots$ Adding $1$ to both sides,we get: $1 + x = 1 + \frac{1}{5} + \frac{1 \cdot 3}{2!} \left(\frac{1}{5}\right)^2 + \frac{1 \cdot 3 \cdot 5}{3!} \left(\frac{1}{5}\right)^3 + \dots$ Using the binomial expansion $(1 - z)^{-n} = 1 + nz + \frac{n(n+1)}{2!} z^2 + \dots$,where $n = 1/2$ and $z = 2/5$: $1 + x = (1 - 2/5)^{-1/2} = (3/5)^{-1/2} = (5/3)^{1/2}$ Squaring both sides: $(1 + x)^2 = 5/3$ $1 + 2x + x^2 = 5/3$ $3 + 6x + 3x^2 = 5$ $3x^2 + 6x = 2$

If $x = \frac{1}{5} + \frac{1 \cdot 3}{5 \cdot 10} + \frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15} + \dots \infty$,then $3x^2 + 6x$ is equal to

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