Summation 1

It's easy to derive from the given recurrance relation that $a_0=\dfrac {3}{2}$ . Let the sum be $S$ . Then

$S=\dfrac {3}{2}+\dfrac {3}{4}+\dfrac {3}{8}+\displaystyle \sum_{n=3}^\infty \dfrac {a_n}{4^n}$

$=\dfrac {21}{8}+2\displaystyle \sum_{n=3}^\infty \dfrac {a_{n-1}}{4^n}+5\displaystyle \sum_{n=3}^\infty \dfrac {a_{n-2}}{4^n}-6\displaystyle \sum_{n=3}^\infty \dfrac {a_{n-3}}{4^n}$

$=\dfrac {21}{8}+2\left (\dfrac {S}{4}-\dfrac {9}{16}\right ) +5\left (\dfrac {S}{16}-\dfrac {5}{32}\right ) -6\dfrac{S}{64}$

$\implies S\left (1-\dfrac {1}{2}-\dfrac {5}{16}+\dfrac {3}{32}\right ) =\dfrac {21}{8}-\dfrac {9}{8}-\dfrac {15}{32}$

$\implies S=\dfrac {11}{3}$

So, $P=11,Q=3$ , and $PQ +P +Q=33+11+3=\boxed {47}$ .

• For the recursion series given above , the characteristic polynomial will be :

  $x³ = 2x² + 5x - 6$

  Thus we get

$x³ - 2x² - 5x + 6 = 0$

• Now we need to find the roots of this polynomial to get the closed form of the recurrence.

  By vieta's formula , we can say that the roots are a factor of 6. And luckily we get that $x = 1$ is a root of this polynomial.

• Furthermore by the factor theorem we know if $x = 1$ is a root , then (x-1) must be a factor or the same polynomial.

  On division we get quotient equal to

$x² - x - 6$

• Now simply use the quadratic formula to get the roots of this polynomial

  $\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

  $\large =\frac{-(-1) \pm \sqrt{(-1)^2-4(1)(-6)}}{2(1)}$

  $\large =\frac{ 1 \pm \sqrt{(1 +24)}}{2}$

  $\large =\frac{ 1 \pm 5}{2}$

  $\large = 3$ or $\large -2$

Finally the roots of the equation are $\large 1,-2,3$

• Now we can write the closed form as

  $a_n = x(1)^n + y(-2)^n + z(3)^n$

  Put the values of $a_1 , a_2 , a_3$ to get three equations for $x,y,z$

  The three equations would be :

  $1.) x - 2y + 3z = 3$

  $2.) x + 4y + 9z = 6$

  $3.) x - 8y + 27z = 18$

  On solving these equations we get $x = 1 , y = -1/10 , z = 3/5$

So the closed for the recursion is: $a_n = (1)^n + \frac{-(-2)^n}{10} + \frac {3(3)^n}{5}$

• Moving towards the summation part

Let S $\large = \sum_{n=0}^\infty \frac{a_n}{4^n}$

$= \sum_{n=0}^\infty \frac{(1)^n + \frac{-(-2)^n}{10} + \frac {3(3)^n}{5}}{4^n}$

$= \sum_{n=0}^\infty \frac{1^n}{4^n} + \frac{-1}{10}\sum_{n=0}^\infty \frac{(-2)^n}{4^n} + \frac{3}{5}\sum_{n=0}^\infty \frac{3^n}{4^n}$

$= \sum_{n=0}^\infty (\frac{1}{4})^n + \frac{-1}{10}\sum_{n=0}^\infty (\frac{-1}{2})^n + \frac{3}{5}\sum_{n=0}^\infty (\frac{3}{4})^n$

$\large = \frac{1}{1-\frac{1}{4}} + (\frac{-1}{10}) \frac{1}{1-\frac{-1}{2}} + (\frac{3}{5} ) \frac{1}{1-\frac{3}{4}}$

$\large = \frac{4}{3} + (\frac{-1}{10}) \frac{2}{3} + (\frac{3}{5} ) \frac{4}{1}$

$\large = \frac{4}{3} - \frac{1}{15} + \frac{12}{5}$

$\large = \frac{20-1+36}{15}$

$\large = \frac{55}{15} = \frac{11}{3}$

Thus $P = 11 , Q = 3$

and the answer is $(11)(3) + 11 + 3$ $\large = 47$