Let $a_n$ be a sequence such that $a_1 = 5$ and $a_{n+1} = a_n + (n - 2)$ for all $n \in N$. Then $a_{51}$ is:

If $S_n = 1^3 + 2^3 + \ldots + n^3$ and $T_n = 1 + 2 + \ldots + n$,then

The sum of the series $1 + (1 + 2) + (1 + 2 + 3) + \dots$ up to $n$ terms is:

Show that $\frac{1 \times 2^{2}+2 \times 3^{2}+\ldots+n \times(n+1)^{2}}{1^{2} \times 2+2^{2} \times 3+\ldots+n^{2} \times(n+1)}=\frac{3 n+5}{3 n+1}$

If the sum of the first $n$ terms of a series is $5n^2 + 2n$,then its second term is

If $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\ldots \infty = \frac{\pi^4}{90}$,$\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+\ldots \infty = \alpha$,and $\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+\ldots \infty = \beta$,then $\frac{\alpha}{\beta}$ is equal to: