Problem source

The sequence $a_{1},a_{2},\ldots,a_{n},\ldots$ forms an arithmetic progression. Establish a formula, involving $n,$ $a_{1}$ and $a_{2}$ for the sum $a_{1}+a_{2}+\cdots+a_{n}$ of the first $n$ terms. 
A sequence $b_{1},b_{2},\ldots,b_{n},\ldots$ is called a \textit{double arithmetic progression} if the sequence of differences
\[
b_{2}-b_{1},b_{3}-b_{2},\ldots,b_{n+1}-b_{n},\ldots
\]
is an arithmetic progression. Establish a formula, involving $n,b_{1},b_{2}$ and $b_{3}$, for the sum $b_{1}+b_{2}+b_{3}+\cdots+b_{n}$ of the first $n$ terms of such a progression. 
A sequence $c_{1},c_{2},\ldots,c_{n},\ldots$ is called a \textit{factorial progression} if $c_{n+1}-c_{n}=n!d$ for some non-zero $d$ and every $n\geqslant1$. Suppose $1,b_{2},b_{3},\ldots$ is a double arithmetic progression, and also that $b_{2},b_{4},b_{6}$ and $220$ are the first four terms in a factorial progression. Find the sum $1+b_{2}+b_{3}+\cdots+b_{n}.$

Solution source

Since the common difference is $a_2 - a_1$ we can find that $a_n = a_1 + (n-1)(a_2-a_1)$, then
\begin{align*}
&& a_1 &&+&& a_2 &&+&& \cdots &&+&& (a_1 + (n-2)(a_2 - a_1) && + && (a_1 + (n-1) (a_2 - a_1)) \\
+ && (a_1 + (n-1) (a_2 - a_1))&&+&& (a_1 + (n-2)(a_2 - a_1)&&+&& \cdots &&+&& a_2 && + && a_1  \\
\hline \\
= && 2a_1 + (n-1)(a_2 - a_1) && + && 2a_1 + (n-1)(a_2 - a_1) && + && \cdots && + 2a_1 + (n-1)(a_2 - a_1) && + 2a_1 + (n-1)(a_2 - a_1) \\
= && n(2a_1 + (n-1) (a_2 - a_1))
\end{align*}

Therefore the sum is $a_1 n + \frac{n(n-1)}{2} (a_2 - a_1)$.

Since $b_n - b_1 = (b_n - b_{n-1}) + (b_{n-1} - b_{n-2}) + \cdots (b_2 - b_1)$, $b_n - b_1 = a_1 (n-1) + \frac{(n-1)(n-2)}{2}(a_2 - a_1) = (b_2-b_1)(n-1) + \frac{(n-1)(n-2)}{2}(b_3 -2b_2 +b_1)$. So $b_n = b_1 + (b_2 - b_1)(n-1) + \frac{(n-1)(n-2)}{2} (b_3 - 2b_2 + b_1)$.

In particular

\begin{align*}
\sum_{i=1}^n b_i &= \sum_{i=1}^n \l b_1 + (b_2 - b_1)(n-1) + \frac{(n-1)(n-2)}{2} (b_3 - 2b_2 + b_1)\r \\
&= nb_1 + (b_2-b_1) \frac{n(n-1)}{2} + \frac{n(n-1)(n-2)}{6}(b_3-2b_2+b_1)
\end{align*}

Let $b_2 - b_1 = x$ and $b_3 - 2b_2+b_1 = y$, then

\begin{align*}
b_4 - b_2 &= d &= &2x + 3y \\
b_6-b_4 &= 2d &=& 2x +(10-3)y \\
&&=&2x + 7y \\
220-b_6&=6d &=& 220-(1 + 5x + 10y) \\
\end{align*}

\begin{align*}
&& 4x + 6y &= 2x + 7y \\
&& 6x+21y &= 219-5x-10y \\
\Rightarrow && 2x - y &= 0 \\
&& 11x + 31y &= 219 \\
\Rightarrow && x &= 3 \\
&& y &= 6
\end{align*}

Therefore the final sum is

\begin{align*}
n + 3 \frac{n(n-1)}{2} + 6 \frac{n(n-1)(n-2)}{6} &= n^3-\frac32n^2+\frac32n
\end{align*}