Questions related to Progressions Find the sum of 
$4+5+7+10+14+19+25+32+40+49+59+70+82$
$+95+109+124+140+157+175+194+214+235+257$.
I want to know the proper method for solving such series questions for these type of long and short series.
 A: As long as there is no known pattern in the series the only way of calculating this sum is by adding each number one by one.
Here you can notice that the difference of the summands is increased by 1 each time, so you could if $a_n$ is the $n$-th summand  $a_{n+1} = a_{n}+n$ and $a_1 = 4$.
So you should try to evaluate $S =\sum\limits_{k=1}^{m} a_k$ ($m=22$ in your case). In order to do that it would be usefull to have an explicit formula for $a_n$:
$$ a_n = 4+ \frac 1 2 (n-1) n = 4+\frac 1 2 n^2 - \frac 1 2 n$$
So $$\begin{align}S &= \sum\limits_{n=1}^m a_n \\ 
&= 4m+ \frac 1 2 \sum\limits_{n=1}^m n^2 - \frac 1 2 \sum\limits_{n=1}^m n \\ 
&= 4m + \frac 1 2 \frac  1 6 m (m+1) (2 m+1) - \frac 1 2 \frac 1 2 m(m+1)\\ 
&= 4m +\frac{1}{12} m (m+1) (2 m+1) - \frac 1 4  m(m+1)\\
&= \frac 1 6 m (m^2+23)\end{align} $$
A: Since the difference of the numbers increases by $1$ each, the numbers in your sum are
$$
a_n=4+\binom{n}{2}
$$
Thus,
$$
\begin{align}
\sum_{n=1}^m\left[4+\binom{n}{2}\right]
&=4m+\binom{m+1}{3}\\
&=4m+\frac{m^3-m}{6}\\[6pt]
&=\frac{m^3+23m}6
\end{align}
$$
