If $2^3+4^3+6^3+\ldots+(2n)^3 = h n^2(n+1)^2$,then $h$ is equal to

Write the first five terms of the sequence whose $n^{th}$ term is $a_{n} = \frac{n}{n+1}$.

If $2^3+4^3+6^3+\ldots+(2n)^3=h n^2(n+1)^2$,then $h$ is equal to

The $n^{th}$ term of the series $\frac{2}{1!} + \frac{7}{2!} + \frac{15}{3!} + \frac{26}{4!} + \dots$ is

The sum of the series $\frac{2^{3}-1^{3}}{1 \times 7}+\frac{4^{3}-3^{3}+2^{3}-1^{3}}{2 \times 11}+\frac{6^{3}-5^{3}+4^{3}-3^{3}+2^{3}-1^{3}}{3 \times 15}+\ldots + \frac{30^{3}-29^{3}+\ldots+2^{3}-1^{3}}{15 \times 63}$ is equal to:

$2^{1/4} \cdot 4^{1/8} \cdot 8^{1/16} \cdot 16^{1/32} \cdots$ is equal to