Evaluate the sum: $\sum\limits_{k = 1}^n {k\left( {1 + \frac{1}{n}} \right)^{k - 1}}$

If $S(x) = (1+x) + 2(1+x)^2 + 3(1+x)^3 + \ldots + 60(1+x)^{60}$,$x \neq 0$,and $(60)^2 S(60) = a(b)^b + b$ where $a, b \in N$,then $(a+b)$ is equal to:

Let $S = 2 + \frac{6}{7} + \frac{12}{7^{2}} + \frac{20}{7^{3}} + \frac{30}{7^{4}} + \ldots$. Then $4S$ is equal to

Suppose $a_1, a_2, 2, a_3, a_4$ are in an arithmetico-geometric progression. If the common ratio of the corresponding geometric progression is $2$ and the sum of all $5$ terms of the arithmetico-geometric progression is $\frac{49}{2}$,then $a_4$ is equal to $...........$.

The sum $\sum\limits_{k = 1}^{20} {k\frac{1}{{{2^k}}}} $ is equal to

$n$th term of the series $1 + \frac{3}{7} + \frac{5}{7^2} + \frac{7}{7^3} + \dots$ is