1 problem found
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 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 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: 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*}