Progressions Of Progrssions

Observe the first terms of all the rows. Let their sum be $\mathbf{S}$ and $n^{\text{th}}$ term be $a_n$ .Then

$\begin{aligned}\mathbf S=&1+2+4+\cdots +a_n\\\mathbf S=&~~~~~~~1+2+\cdots+a_{n-1}+a_n\end{aligned}$

Substracting and rearranging for $a_n$ , we get:

$a_n=1+(1+2+3+\cdots (n-1)\text{terms})\\=\dfrac{n^2+n+1}{2}$

We can observe that $n^{\text{th}}$ row contains $n$ terms, it's first term is $a_n$ and common difference is $1$ . Thus sum of $n^{\text{th}}$ row (An AP) is:

$a_n+a_{n+1}+a_{n+2}\cdots(n\text{ terms})~~\\=\dfrac{n}{2}\left(2(\dfrac{n^2+n+1}{2})+(n-1)\right)~~$

For $n=2016$ , this is:

$\large 1008(2016^2+1)$

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For calculating its last $3$ digits, we observe last $3$ digits of $2016^2$ (same as that of $016^2$ ) are $256$ plus one is $257$ multiplied by 8 we get its last three digits as 056

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Note :- There is no need to calculate the whole number..

We note that $n^{th}$ row has $n$ terms and its last term is the $n^{th}$ triangular number $T_n = \dfrac{n(n+1)}{2}$ .

Let $a$ and $l$ be the first and last terms of $2016^{th}$ row respectively, then the sum $S$ is given by:

$\begin{aligned} S & = \frac{2016(a-l)}{2} \quad \quad \small \color{#3D99F6}{\text{Note that }a = T_{2015}+1 \text{ and }l = T_{2016}} \\ & = 1008 \left(T_{2015}+1 + T_{2016} \right) \\ & = 1008 \left( \frac{ 2015 (2016)} {2} + 1 + \frac{2016(2017)}{2} \right) \\ & = 1008 \left(2031120 + 1 + 2033136 \right) \\ & = 4096771\boxed{056} \end{aligned}$

The solution is a little bit lengthy.....

The $n^{th}$ row has $n$ numbers in it.

Therefore the total numbers till the $2015^{th}$ row $=~\large\frac{2015 \times 2016}{2}$ $=~2031120$

The first number in $2016^{th}$ row $=~2031121$

The last number in $2016^{th}$ row $=~2031121~+~2016~=~2033136$

Therefore sum of numbers till the $2015^{th}$ row $=~\large\frac{2031120 \times 20311121}{2}$ $=~2062725242760$

Therefore sum of numbers till the $2016^{th}$ row $=~\large\frac{2033136 \times 2033137}{2}$ $=~2066822013816$

Thus the sum of numbers in the $2016^{th}$ row = $2066822013816~-~2062725242760~=~ 4096771056$

Hence the last three digits are $\boxed{056}$

It's a finite difference sequence.