Question

#5. Given a sequence fa pao, a , a, , define the sequence lbne bobt b2, by 田 b.-ao + a1 +a2 + + ak If a (x) is the generating function for (an) and b(x) is the generating function for tbnj, then show (1-x)a. This lets us write b)4 #6. Let(anpao, ai, a2, given by a,-0, ai-1, an+2-an+1 + an be the sequence of Fibonacci numbers; recall that the ordinary generating function for the Fibonacci numbers is a(x)Let (bn] be the sequence defined by bk-: ao + ai + + ak as in problem #5. Thus, bk is the sum of the Fibonacci numbers up to ak. The sequence {bn} begins 0,1,2,4,7,12, From #5, the generating function for {m} is a-x-x2)(1-x) Use partial fractions decomposition to find constants r,s,t such that sx +t = t (1-x-x2)(1-x) 1-x 1-x-x2 Explain how your results can be used to derive the identity b an+1 1. ENG 11:34 PM 1/22/2019

solve 6 please . step by step.

Answer 1

Solution:

we have,

$\frac{x}{(1-x-x^2)(1-x)}=\frac{r}{1-x}+\frac{sx+t}{1-x-x^2}$

$=>\frac{x}{(1-x-x^2)(1-x)}=\frac{r\left ( 1-x-x^2 \right )+\left (sx+t \right )\left ( 1-x \right )}{\left (1-x-x^2 \right )\left ( 1-x \right )}$

$=>\frac{x}{(1-x-x^2)(1-x)}=\frac{r-rx-rx^2 +sx-sx^2+t-tx }{\left (1-x-x^2 \right )\left ( 1-x \right )}$

$=>\frac{x}{(1-x-x^2)(1-x)}=\frac{r+t+\left (s-r-t \right )x-\left (r+s \right )x^2 }{\left (1-x-x^2 \right )\left ( 1-x \right )}$

Now comparing the numerator with the coefficients of the corresponding powers of x we get,

$=>r+t=0----(1)$

$=>s-r-t=1----(2)$

$=>-(r+s)=0----(3)$

from (1),(2),(3) we get that,

$=>r=-1,s=t=1$

Therefore we can decompose bn into

$=>\left \{ b_n \right \}=\frac{-1}{1-x}+\frac{x+1}{1-x-x^2}$