Let S = 2 + (6)/(7) + 12 7 ^ 2 + 20 7 ^ 3 + 30 7 ^ 4 + . . then 4S...

$S = 2 + \dfrac{6}{7} + \dfrac{12}{{7}^{2}} + \dfrac{20}{{7}^{3}} + ⋯ . . . \left(i\right)$

$\dfrac{S}{7} = 0 + \dfrac{2}{7} + \dfrac{6}{{7}^{2}} + \dfrac{12}{{7}^{3}} + \dfrac{20}{{7}^{4}} + ⋯ . . . \left(ii\right)$

On subtracting the equation \left(ii\right)$from equation \left(i\right)$, we get

$\dfrac{6S}{7} = 2 + \dfrac{4}{7} + \dfrac{6}{{7}^{2}} + \dfrac{8}{{7}^{3}} + ⋯ . . . \left(iii\right)$

$\dfrac{6S}{49} = 0 + \dfrac{2}{7} + \dfrac{4}{{7}^{2}} + \dfrac{6}{{7}^{3}} + ⋯ . . . \left(iv\right)$

On subtracting the equation \left(iv\right)$from equation$\left(iii\right)$, we get</p><p>\left(\dfrac{6}{7} - \dfrac{6}{49}\right)S = 2 + \dfrac{2}{7} + \dfrac{2}{{7}^{2}} + \dfrac{2}{{7}^{3}} + ⋯$

$\dfrac{36}{49}S = 2\left(1 + \dfrac{1}{7} + \dfrac{1}{{7}^{2}} + ⋯ \infty\right)$

$⇒ S = 2 \cdot \dfrac{1}{1 - \dfrac{1}{7}} \cdot \dfrac{49}{36}$

$⇒ 4S = {\left(\dfrac{7}{3}\right)}^{3}$