Muhammad's Recurrence

The first step is to decompose the summand by partial fractions. $\frac{a_{i-1}}{a_{i}^2 - a_{i-1}^2} = \frac{a_{i-1}}{(a_{i} - a_{i-1})(a_{i}+a_{i-1})}$

$\frac{a_{i-1}}{a_{i}^2 - a_{i-1}^2} = \frac{1}{2(a_{i}-a_{i-1})} - \frac{1}{2(a_{i}+a_{i-1})}$

Now, since $a_{n+1} = 2a_{n} + a_{n-1}$ , we get $a_{n+1} - a_{n} = a_{n} + a_{n-1}$ . This simplifies our summand to:

$\frac{1}{2(a_{i}-a_{i-1})} - \frac{1}{2(a_{i+1}-a_{i})}$ , which telescopes, so when added, the sum will be only the first fraction.

$S = \frac{1}{2(a_1 - a_0)} = \frac{1}{2(213-1)} = \frac{1}{424}$

Giving the answer of $\frac{1}{S} = 424$

We have: $\frac {2a_{i-1}}{a^2_i-a^2_{i-1}} = \frac {(a_i+a_{i-1})-(a_i-a_{i-1})}{(a_i+a_{i-1})(a_i-a_{i-1})} = \frac {1}{a_i-a_{i-1}} - \frac {1}{a_i+a_{i-1}} = \frac {1}{a_{i-1}+a_{i-2}} - \frac {1}{a_i+a_{i-1}}$ for $i \geq 2$ , thus $2S = \frac {1}{a_1-a_0} - \frac {1}{a_1+a_0} + \frac {1}{a_1+a_0} - \frac {1}{a_2-a_1} + \ldots = \frac {1}{a_1-a_0} = \frac {1}{212}$ , and thus $\frac 1S = 424$

Consider that

a {i+1}-a {i}=a {i}+a {i-1}

and

\frac{a {i-1}}{a {i}^2-a {i-1}^2}=\frac{1}{2}(\frac{1}{a {i}-a {i-1}}-\frac{1}{a {i}+a_{i-1}})

for every i \in N. Hence,we have

S=\frac{1}{2}(\frac{1}{a {1}-a {0}}-\frac{1}{a {1}+a {0}})+\frac{1}{2}(\frac{1}{a {2}-a {1}}-\frac{1}{a {2}+a {1}})+\frac{1}{2}(\frac{1}{a {3}-a {2}}-\frac{1}{a {3}+a {2}})+\ldots

which equivalent with

S=\frac{1}{2}(\frac{1}{a {1}-a {0}}-\frac{1}{a {1}+a {0}})+\frac{1}{2}(\frac{1}{a {1}+a {0}}-\frac{1}{a {2}+a {1}})+\frac{1}{2}(\frac{1}{a {2}+a {1}}-\frac{1}{a {3}+a {2}})+\ldots

After several process, we get

S=\frac{1}{2}(\frac{1}{a {1}-a {0}})

which equivalent with

S=\frac{1}{2}(\frac{1}{213-1})

so that S=\frac{1}{424}. So, the value of \frac{1}{S} is 424.

Note that

$\displaystyle\frac{1}{a_i-a_{i-1}}-\frac{1}{a_{i+1}-a_i}=\frac{a_{i+1}-a_i-a_i+a_{i-1}}{(a_i-a_{i-1})(a_{i+1}-a_i)}$

$=\displaystyle\frac{a_{i+1}-2a_i+a_{i-1}}{(a_i-a_{i-1})(a_i+a_{i-1})}=\frac{2a_{i-1}}{(a_i-a_{i-1})(a_i+a_{i-1})}=\frac{2a_{i-1}}{a_i^2-a_{i-1}^2}$

since $a_{i+1}=2a_i+a_{i-1}$ . Hence,

$S=\frac{1}{2}\displaystyle\sum^{\infty}_{i=1}\frac{2a_{i-1}}{a_i^2-a_{i-1}^2}=\frac{1}{2}\displaystyle\sum^{\infty}_{i=1}\left(\frac{1}{a_i-a_{i-1}}-\frac{1}{a_{i+1}-a_i}\right)$

The summation telescopes to

$\displaystyle S=\frac{1}{2}\cdot\frac{1}{a_1-a_0}=\frac{1}{424},$

so $\frac{1}{S}=\boxed{424}$ .

$S = \sum_{i=1}^\infty \frac{a_{i-1}}{(a_i + a_{i-1})(a_i - a_{i-1})}$

$= \frac{1}{2} \sum_{i=1}^\infty (\frac{1}{a_i - a_{i-1}} - \frac{1}{a_i + a_{i-1}})$ ,

which is basically a telescoping series problem, as according to the given recurrence relation,

$a_{i+1} - a_i = a_i + a_{i-1}$ ,

and so the second term of this expression is cancelled by the first term of the next expression as $i$ increments & so on.

The term remaining is the first one:

$S = \frac{1}{2} \frac{1}{a_1 - a_0} = \frac{1}{424}$ .

$\frac{1}{S} = 424$ , as required.

Using partial fractions, $\frac{a_{i-1}}{a_i^2-a_{i-1}^2} = \frac{1}{2}\left(\frac{1}{a_i-a_{i-1}}-\frac{1}{a_i+a_{i-1}}\right).$

By the recurrence relation, $a_i+a_{i-1}=a_{i+1}-a_i$ , implying $\begin{aligned} S &= \displaystyle\sum_{i=1}^\infty \frac{1}{2}\left(\frac{1}{a_i-a_{i-1}} - \frac{1}{a_{i+1}-a_i}\right) \\ & = \frac{1}{2(a_1 - a_0)} = \frac{1}{424}. \end{aligned}$

Therefore, $\frac{1}{S}= 424$ .

Note: Here the series is calculated using the fact that it is telescoping. Formally, a partial sum of this series, after all terms in the middle cancel out, becomes $\frac{1}{a_1-a_0}-\frac{1}{a_{i+1}-a_{i}}.$ We can easily see that $a_i \rightarrow \infty$ , hence $\frac{1}{a_{i+1}-a_{i}} =\frac{1}{a_{i} + a_{i-1}} \rightarrow 0,$ so the sum converges to $\frac{1}{a_1-a_0}$ .

let know that the first is 1/212

ratio:1/2

so, a/(1-r) (1/212)/2 =1/424

s=1/424 1/s=424

$1/((213)^2 - 1^2) + 213/((2213 + 1)^2 - 213^2) + (2213 + 1)/((5213 + 2)^2 - (2213 + 1)^2) + ...$ $= 1/(213214) + 213/(214(3213 + 1)) + (2213 + 1)/((3213 + 1)(7213 + 3)) + ...$ $= (1/2)(1/(212) - 1/(214)) + (1/2)(1/(214) - 1/(3213 + 1)) + (1/2)(1/(3213 + 1) - 1/(7213 + 3)) + ...$ $= (1/2) (1/(212)) - (1/2)(1/214) + (1/2)(1/214) - (1/2)(1/(3213 + 1) + (1/2)(1/(3*213 + 1)) - ...$ then we get S = (1/2)(1/212) = 1/424