The first term of the ${11^{th}}$ group in the following sequence of groups $(1), (2, 3, 4), (5, 6, 7, 8, 9), \dots$ is

The sum of the infinite series $1+\frac{2}{3}+\frac{7}{3^{2}}+\frac{12}{3^{3}}+\frac{17}{3^{4}}+\frac{22}{3^{5}}+\ldots$ is equal to

The sum of the first $n$ terms of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \dots$ is $\frac{n(n + 1)^2}{2}$ when $n$ is even. When $n$ is odd,the sum is:

If $\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^i {\sum\limits_{k = 1}^j {1 = 560} } } $,then the value of $n$ is:

If the sum of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + \dots + 2 \cdot (n-1)^2 + n^2$ (when $n$ is odd) is to be determined,given that for even $n$,the sum is $\frac{n(n+1)^2}{2}$,find the sum when $n$ is odd.

The $9^{th}$ term of the sequence $27, 9, 5\frac{2}{5}, 3\frac{6}{7}, \dots$ is $.....$