If $\sum_{k=1}^n (2k-1)a_k = n(n+1)(4n-1)$, find $a_n$

Question

I'm not getting any idea what to do with this problem. If given that $\displaystyle\sum\limits_{k=1}^n (2k-1)a_k = n(n+1)(4n-1)$ then how can we find out $a_n$.

I need some hints to start.

Answer 1

Finding a solution

One way is to look for a solution of the form $a_n=\alpha n + \beta$.

Then $$\begin{split} n(n+1)(4n-1)&=\sum_{k=1}^n (2k-1)a_k \\ &=\sum_{k=1}^n (2k-1)(\alpha k+\beta)\\ &= 2\alpha \left(\sum_{k=1}^n k^2\right)+(2\beta-\alpha)\left(\sum_{k=1}^n k\right) -\beta \left(\sum_{k=1}^n 1\right)\\ &=2\alpha\frac{n(n+1)(2n+1)}{6}+(2\beta-\alpha)\frac{n(n+1)}{2}-\beta n \end{split}$$ and by expanding the right-hand side, you can identify the value of $\alpha$ and $\beta$.

This is the only solution

This finds a solution, but doesn't show this is the only solution. But we can prove it is.

Note that if $a=(a_1,a_2,\dots, a_n)^T$ and $b=(6, 42, \dots, n(n+1)(4n-1))^T$, then any solution $a$ verifies $Ma=b$ where $$M=\left( \begin{array}{ccccc} 1 & 0 & 0&\dots &0\\ 1&3& 0&\dots&0\\ \vdots\\ 1&3&5&\dots&n(n+1)(4n-1) \end{array}\right)$$ This matrix is obviously invertible, so there is only one solution to $Ma=b$.

Addendum Note that the inverse of $M$ can be explicitly computed. Indeed, $M=PD$ where $$P=\left( \begin{array}{ccccc} 1 & 0 & 0&\dots &0\\ 1&1& 0&\dots&0\\ \vdots&\vdots&\ddots\\ 1&1&1&\dots&1 \end{array}\right)$$ and $D$ is the diagonal matrix $$D=\left( \begin{array}{ccccc} 1 \\ &3& \\ &&\ddots\\ &&&n(n+1)(4n-1) \end{array}\right)$$ $D$ is easy to invert, and $$P^{-1}=\left( \begin{array}{ccccc} 1 & 0 \dots \\ -1&\ddots& \ddots&\\ 0&\ddots&\ddots\\ \vdots&\ddots&-1&1\\ \end{array}\right)$$

Answer 2

Setting $n=1$ we get $a_1 = 6$.

Call $b_n = \sum\limits_{k=1}^n (2k-1)a_k = n(n+1)(4n-1)$.

Then, for $n>1$, on the one hand, we have $b_n - b_{n-1} = (2n-1) a_n$, on the other hand we have $b_n - b_{n-1} = n(n+1)(4n-1) - (n-1)(n)(4(n-1)-1) = (2n-1) 6 n$.

Cancel $2n-1$ to get $a_n = 6n$.