# I want to solve the equation b n </msub> = b <mrow class="MJX-TeXAtom-OR

Carolyn Beck 2022-06-29 Answered
I want to solve the equation
${b}_{n}={b}_{n-1}+2{b}_{n-2}+n+1$
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## Answers (1)

pyphekam
Answered 2022-06-30 Author has 27 answers
${b}_{n}={b}_{n-1}+2{b}_{n-2}+n+1$
The guess is
${b}_{n}=cn+d$
$cn+d=c\left(n-1\right)+d+2\left(c\left(n-2\right)+d\right)+n+1$
$-c+2cn-4c+2d+n+1=0$
$-5c+1+2d+n\left(2c+1\right)=0$
$2c+1=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}c=-\frac{1}{2}$
$\frac{5}{2}+1+2d=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}d=-\frac{7}{4}$
So that:
${b}_{n}={c}_{1}\left(-1{\right)}^{n}+{c}_{2}{2}^{n}-\frac{n}{2}-\frac{7}{4}$
For ${b}_{1}={b}_{0}=1$
${b}_{0}={c}_{1}+{c}_{2}-\frac{7}{4}$
${c}_{1}+{c}_{2}=\frac{11}{4}$
Then
${b}_{1}=-{c}_{1}+2{c}_{2}-\frac{1}{2}-\frac{7}{4}$
$-{c}_{1}+2{c}_{2}=\frac{13}{4}$
Then we have:
${c}_{1}=\frac{3}{4},{c}_{2}=2$
${b}_{n}={c}_{1}\left(-1{\right)}^{n}+{c}_{2}\left({2}^{n}\right)-\frac{n}{2}-\frac{7}{4}$
${b}_{n}=\frac{3}{4}\left(-1{\right)}^{n}+2\left({2}^{n}\right)-\frac{n}{2}-\frac{7}{4}$
${b}_{2}=\frac{3}{4}+2\left({2}^{2}\right)-\frac{2}{2}-\frac{7}{4}$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{b}_{2}=6$
As expected.
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