auto23652im

2022-07-14

The sequence $\left(b{\right)}_{n\ge 0}$ shall satisfy
${b}_{n}=\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right){a}_{k}$

Ordettyreomqu

Expert

For every $n\ge 0$, one has
$\begin{array}{rl}{b}_{n}& =\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right){a}_{k}\\ & =\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right)\frac{6\cdot {3}^{k}+\left(-1{\right)}^{k+1}{2}^{k}}{5}\\ & =\frac{6}{5}\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right){3}^{k}-\frac{1}{5}\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right)\left(-2{\right)}^{k}\\ & =\frac{6}{5}\left(3+1{\right)}^{n}-\frac{1}{5}\left(1-2{\right)}^{n}\\ & =\frac{6\cdot {4}^{n}-\left(-1{\right)}^{n}}{5}\end{array}$
which satisfies the recursion
$\overline{){b}_{n+2}=3{b}_{n+1}+4{b}_{n}}$

Do you have a similar question?