For $\frac{2^2+4^2+6^2+\ldots+(2n)^2}{1^2+3^2+5^2+\ldots+(2n-1)^2}$ to exceed $1.01$,the maximum value of $n$ is

$\frac{10001 \times 100 !}{2 \times 1 !+5 \times 2 !+10 \times 3 !+\ldots+10001 \times 100 !}=$

If the sum of the first $11$ terms of the series ${\left( {1\frac{4}{7}} \right)^2} + {\left( {1\frac{5}{7}} \right)^2} + {\left( {1\frac{6}{7}} \right)^2} + {2^2} + {\left( {2\frac{1}{7}} \right)^2} + \dots$ is $\frac{11}{7}\lambda$,then $\lambda$ is equal to:

Let $a_{n}$ be the $n^{\text{th}}$ term of the series $5+8+14+23+35+50+\ldots$ and $S_{n}=\sum_{k=1}^{n} a_{k}$. Then $S_{30}-a_{40}$ is equal to

$\sum_{k=1}^{\infty} \sum_{r=0}^k \frac{1}{3^k} \binom{k}{r}$ is equal to

The sum of the series $1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + \dots$ to $n$ terms is: