Such Years, Such Summations, Much Wow

First see what $a_i=\displaystyle\sum_{j=i+1}^{2016} a_j$ exactly means. It means that the term $a_i$ equals the sum of the terms successive to $a_i$ in the sequence. So, if we let $S=\displaystyle\sum_{i=1}^{2016} a_i=S$ we get the following relations:

$\begin{aligned} S&= 2a_1 \\ S&=a_1+2a_2 \\ S&=a_1+a_2+2a_3 \\ \dots\dots \\ \dots\dots \\ S&=a_1+a_2+a_3+\dots+a_{2013}+2a_{2014} \\ S&=a_1+a_2+a_3+\dots+a_{2014}+2a_{2015} \end{aligned}$

Let us sum all the $2015$ equations of $s$ written above and we arrive at this equation:

$2015S=2016a_1+2015a_2+2014a_3 + \dots + 4a_{2013}+3a_{2014}+2a_{2015}$

Note that the RHS above is nothing but $\displaystyle\sum_{i=0}^{2014} [(2016-i)a_{i+1}]$ that equals $2015$ which is given.

So we have $2015S=2015 \Rightarrow S=1$

Substituting $S=1$ in the above $2015$ equations, we can individually get all the terms in the sequence $<a_i>$ as $\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8},\dots, \dfrac{1}{2^{2015}},\dfrac{1}{2^{2015}}$ . Thus, we have $a_i=\dfrac{1}{2^i}$ .

Isn't this astonishing? So lets take reciprocal:: $\dfrac{1}{a_i}=2^i$ and we need to just sum it using geometric progression sum to get the answer:

$\displaystyle\sum_{i=1}^{2015} 2^i + 2^{2015}=\dfrac{2(2^{2015}-1)}{2-1}+2^{2015}=2^{2015}-2+2^{2015}=3\times 2^{2015} - 2$

So $a+b+c-1=2+2015+3-1=\huge\boxed{2019}$

credits to @Saurabh Chaturvedi @Billy Sugiarto for correct my solution!

To begin with, let's prove the following statements.

• $a_i = a_{2016} \times 2^{2015 - i}$ for $1 \leq i \leq 2015$ $(1)$

For $i = 2015$ we have $a_{2015} = \displaystyle \sum_{j = 2016}^{2016}{a_j} = a_{2016} = a_{2016} \times 2^0$ . Let's assume that the statement holds for $i = k$ . Then $a_{k-1} = \displaystyle \sum_{j = k}^{2016}{a_j} = a_{2016} \times 2^{2015 - k} + a_{2016} \times 2^{2015- (k + 1)} + \ldots + a_{2016} \times 2 + a_{2016} + a_{2016} = a_{2016} \times \big(1 + (1 +\\+ 2 + \ldots + 2^{2015-k})\big) = a_{2016} \times 2^{2015-(k - 1)}.$

• $\displaystyle \sum_{i = 1}^{x}{i \times 2^{x - i}} = 2^{x + 1} - (x + 2)$ $\forall x \in N$ $(2)$

For $x = 1$ we have $\displaystyle \sum_{i=1}^{1}{i \times 2^{1 - i}} = 1 = 2^{1 + 1} - (1 + 2)$ . Let's assume that the statement holds for $x = k$ . Then for $x = k + 1$ we have

$\displaystyle \sum_{i = 1}^{x}{i \times 2^{x - i}} = \displaystyle \sum_{i = 1}^{k + 1}{i \times 2^{k + 1 - i}} = 2 \times \displaystyle \sum_{i = 1}^{k}{i \times 2^{k - i}} + (k + 1) \times 2^0 = 2 \times \big(2^{k + 1} - (k + 2)\big) + k + 1 =\\= 2^{(k + 1) + 1} - ((k + 1) + 2).$

Applying $(1)$ and $(2)$ , we have

$2015 = \displaystyle \sum_{i = 0}^{2014}\big((2016 - i) \times a_{i + 1}\big) = \displaystyle \sum_{i = 0}^{2014}\big((2016 - i) \times a_{2016} \times 2^{2014 - i}\big) = a_{2016} \times (2016 \times \displaystyle \sum_{i = 0}^{2014}2^{2014 - i} - \displaystyle \sum_{i = 0}^{2014}i \times \\ \times 2^{2014 - i}) = a_{2016} \times \big(2016 \times (2^{2015} - 1) - (2^{2015} - 2016)\big) = a_{2016} \times 2015 \times 2^{2015}.$

Hence, $a_{2016} = 2^{-2015}.$

FInally,

$\displaystyle \sum_{i = 1}^{2016} \frac 1{a_i} = \frac 1{a_{2016}} \times \big(1 + (1+ 2^{-1} + \ldots + 2^{-2014})\big) = 2^{2015} \times \big(1 + \frac {2^{-2015} - 1}{2^{-1} - 1}\big) = 2^{2015} \times \big(1 + 2 \times (1 - 2^{-2015})\big) =\\= 2^{2015} \times (3 - 2^{-2014}) = 2^{2015} \times 3 - 2.$

Therefore, $a + b + c - 1 = 2 + 2015 + 3 - 1 = \boxed{2019}.$