Get Ready for 2018!

The first few terms of this series are -

0 + 1 + 2 + 2 +......

Therefore, ${T}_{n} = n({2}^{n-1}$ [There are better ways to explain this too which can be found easily]

First we would find a way to find sum of all the terms till n.

$S = 1(1) + 2(2) + 3({2}^{2}) + .... + n({2}^{n-1})$

$2S = 1(2) + 2({2}^{2}) + 3({2}^{3}) + ..... + (n-1)({2}^{n-1}) + n({2}^{n})$

$S - 2S = 1 + 2 + {2}^{2} + {2}^{3} + ...... + {2}^{n-1} - n({2}^{n}$

Now, $(1 + 2 + {2}^{2} + {2}^{3} + ...... + {2}^{n-1})$ is in GP

$-S = {2}^{n} - 1 - n({2}^{n})$

$S = {2}^{n}(n-1) + 1$

Now, for S < 2018, n = 8 which can be found easily.

Hence,

${S}_{8} + \left\lceil { log }_{ 2 }{ (k }_{ 1 }) \right\rceil \quad +\quad \left\lceil { log }_{ 2 }({ k }_{ 2 }) \right\rceil \quad +.....\quad \left\lceil { log }_{ 2 }{ (k }_{ n }) \right\rceil = 2018$

where $\quad \left\lceil { log }_{ 2 }{ (k }_{ n }) \right\rceil$ is same as $\quad \left\lceil { log }_{ 2 }n) \right\rceil$

Now, P = $\left\lceil { log }_{ 2 }{ (k }_{ 1 }) \right\rceil \quad +\quad \left\lceil { log }_{ 2 }({ k }_{ 2 }) \right\rceil \quad +.....\quad \left\lceil { log }_{ 2 }{ (k }_{ n }) \right\rceil$

hence,

1793 + P = 2018

P = 225

Now, each term of P will be equal to 9 as last term of ${S}_{8}$ is $\left\lceil { log }_{ 2 }256 \right\rceil$

So, P = 9x and P has x terms

9x = 225

x = 25

Therefore, n = 256 + 25 = $\boxed{281}$

$\text{We may observe that the number of terms that give us} \left\lceil { Log_2 X } \right\rceil =N ~ is ~ 2^{N- 1},\\ when ~ X ~ takes ~ the ~ values ~ from ~( 2^{N - 1}+1) ~~ to ~ 2^N.\\ \therefore\quad \quad we \quad have :-\quad \quad \quad \\ For ~ X\quad \quad \quad N\quad \quad \quad terms \quad \quad \quad N*terms \quad \quad \quad Total\\ \quad \quad 1\quad \quad \quad 0\quad \quad \quad \quad 0\quad \quad \quad \quad \quad \quad 0\quad \quad \quad \quad \quad \quad \quad 0\\ \quad \quad ~2 \quad \quad \quad 1 \quad \quad \quad \quad 2^0 \quad \quad \quad \quad \quad 1 \quad \quad \quad \quad \quad \quad \quad \quad 1 \\ \quad ~3~to~4 \quad \quad \quad 2 \quad \quad \quad \quad 2^1 \quad \quad \quad \quad \quad 4 \quad \quad \quad \quad \quad \quad \quad \quad 5 \\ \quad ~5~to~8 \quad \quad \quad 3 \quad \quad \quad \quad 2^2 \quad \quad \quad \quad \quad 12 \quad \quad \quad \quad \quad \quad \quad \quad 17 \\ \quad ~ 9~to~16 \quad \quad \quad 4 \quad \quad \quad \quad 2^3 \quad \quad \quad \quad \quad 32 \quad \quad \quad \quad \quad \quad~~~ \quad 49\\ \quad 17~to~32 \quad \quad \quad 5 \quad \quad \quad \quad 2^4 \quad \quad \quad \quad \quad 80 \quad \quad \quad \quad \quad \quad \quad \quad 129 \\ \quad 33~to~64 \quad \quad \quad 6 \quad \quad \quad \quad 2^5 \quad \quad \quad \quad \quad 192 \quad \quad \quad \quad \quad \quad \quad \quad 321 \\ \quad 65~to~128 \quad \quad \quad 7 \quad \quad \quad \quad 2^6 \quad \quad \quad \quad \quad 448 \quad \quad \quad \quad \quad \quad \quad \quad 769 \\ \quad 129~to~256 \quad \quad \quad 8 \quad \quad \quad \quad 2^7 \quad \quad \quad \quad \quad 1024 \quad \quad \quad \quad \quad \quad \quad \quad 1793 \\ \text{. We are short by 2018 - 1793 =225.}\\ \text{Next step has 256 terms of 9s. So we are in a zone where a few terms of 9 would suffice.}\\ \text{Our X is now 256. we need } \dfrac {225} 9=25~ more ~terms. ~So~~ n= 256+25=\huge \color{#D61F06}{281}$

This is a Arithmetic-Geometric Progression. We may use the formula given in Wiki. The series is (n_a=n*2^{n-1}. Kartik Sharma has used the same method given by Wiki.