Question

If the sum of $n$ terms of the series $5+7+13+31+85+\dots$ is $\frac{1}{2}(a^n+bn+c)$, then

• A. $b+5c=a$
• B. $b+4c=a$
• C. $\frac{b}{a+c}=4$
• D. $\frac{b}{a+c}=2$

Answer 1

Answer: (A), (C)

$\frac{b}{a+c}=4$

Let
$S_n=5+7+13+31+85+\dots +T_n\cdots \left(i\right)S_n=5+7+13+31+85+\dots T_{n-1}+T_n\cdots \left(ii\right)$
from $\left(i\right)-\left(ii\right)$ we get
$0=5+[2+6+18+54+\cdots +(T_n-T_{n-1})]-T_n{T}_n=5+2\frac{(3^{n-1}-1)}{2}=4+3^{n-1}$
Now $S_n=\sum T_n$
$=\sum 4+\sum 3^{n-1}=4n+[1+3+3^2+3^3+\cdots +3^{n-1}]=4n+\frac{3^n-1}{2}=\frac{1}{2}(3^n+8n-1)$
So comparing it with $\frac{1}{2}(a^n+bn+c)$ we have
$a=3,b=8,c=-1$