Animated Solution for Mathematics - Sequence and Series: Let $S=2+\frac{6}{7}+\frac{12}{7^2}+\frac{20}{7^3}+\frac{30}{7^4}+\dots$. then $4S$ is equal to

The given infinite series is $S=2+\frac{6}{7}+\frac{12}{7^2}+\frac{20}{7^3}+\frac{30}{7^4}+\dots$.

At first glance, the denominators are powers of $7$, suggesting a Geometric Progression. However, the numerators $2,6,12,20,30,\dots$ follow a quadratic pattern, identifying this as a second-order Arithmetico-Geometric Progression (AGP).

The First Shift

To simplify, we employ the 'Shift-and-Subtract' technique. We multiply the entire series $S$ by the common ratio of the geometric component, which is $\frac{1}{7}$.

Subtracting $\frac{S}{7}$ from $S$ yields $\frac{6S}{7}$:

Let us define this new series as $S_1=\frac{6S}{7}$. The numerators are now linear ($4,6,8,\dots$), confirming the reduction of the series order.

The Second Reduction

We repeat the 'Shift-and-Subtract' process on $S_1$ to reduce the linear numerators to a constant. Multiply $S_1$ by $\frac{1}{7}$:

Subtracting these equations results in:

The Grand Finale

The series now consists of a constant term plus an infinite geometric progression. We can express this as:

The sum of the infinite GP inside the parentheses is $\frac{a}{1-r}$, where $a=\frac{1}{7}$ and $r=\frac{1}{7}$:

Substituting this back into our equation:

Solving for $S_1$, we find $S_1=\frac{7}{3}\cdot \frac{7}{6}=\frac{49}{18}$. Since $S_1=\frac{6S}{7}$, we solve for $S$:

Finally, calculating $4S$ as requested: