$\sum\limits_{m = 1}^n {{m^2}}$ is equal to

The terms of a $G.P.$ are positive. If each term is equal to the sum of two terms that follow it,then the common ratio is

Let $a, b, c, d$ and $p$ be any non-zero distinct real numbers such that $(a^{2}+b^{2}+c^{2}) p^{2}-2(ab+bc+cd) p+(b^{2}+c^{2}+d^{2})=0$. Then:

The sum of the integers from $1$ to $100$ which are not divisible by $3$ or $5$ is

Let ${S_n} = \frac{1}{{{1^3}}} + \frac{{1 + 2}}{{{1^3} + {2^3}}} + \frac{{1 + 2 + 3}}{{{1^3} + {2^3} + {3^3}}} + \dots + \frac{{1 + 2 + \dots + n}}{{{1^3} + {2^3} + \dots + {n^3}}}$. If $100 S_n = n$,then $n$ is equal to:

The sum of the series $1 + 2 \times 3 + 3 \times 5 + 4 \times 7 + \dots$ up to the $11^{th}$ term is