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I need to evaluate the following sum:

$$102+105+108+\ldots+999$$

I've just been introduced to rule of sum, rule of product, double counting, and complement for a counting unit in my math class. I'm a little lost as to how I should solve this using those theories.

I've come up with two ways, but I still don't understand comparing it to my teachers solutions of similar examples.


Method 1:

$$\begin{align}&102+105+108+\ldots+999 \\ = \quad&(3\times34)+(3\times35)+(3\times36)+\ldots+(3\times333) \\ = \quad&3(34+35+36+\ldots+333) \\ = \quad&3(333\times334)/2 = 166,833 \end{align}$$ $$\boxed{ \ \text{Sum} = 166,833 \ }$$


Method 2:

$$\begin{align} 999-102 &= 897 \\ 897\div 3 &= 299\tag{dividing by $3$; the sum increases by $3$}\end{align}$$ Add $3$, $299$ times to reach $999$.

Plus the first term which results in $300$ terms to sum. So, $$(102+999/2)\times300 = 165,150$$ $$\boxed{ \ \text{Sum} = 165,150 \ }$$


I understand how to solve $1+2+3+\ldots+100$. $$\begin{align} \text{Sum} &= 1+2+3+\ldots+100 \\ &= 100+99+98+\ldots+1 \end{align}$$ $$\Downarrow$$ $$\begin{align} 2\times\text{Sum} &= 101+101+101+\ldots+101\tag{adding sums together} \\ &= \frac{100\times101}{2} \tag*{$\left(\begin{align} &\text{multiplying by $100$;} \\ &\text{there are $100$ terms.}\end{align}\right)$} \end{align}$$ $$\boxed{ \ \text{Sum} = 5050 \ }$$

What am I missing to solve sums of different consecutive numbers?

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  • $\begingroup$ I edited but I could not understand the end of method $2$. At that part, I just added dollar signs around what you wrote. $\endgroup$ – Mr Pie May 10 '18 at 3:29
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You got the right idea but made a mistake. Note that $$ S_n = \sum_{k=1}^n k = \frac{n(n+1)}{2}. $$ You have $$ \sum_{k=34}^{333} 3k = 3 \sum_{k=34}^{333} k = 3 (S_{333} - S_{33}) = 3 \times \frac{333 \cdot 334 - 33 \cdot 34}{2} = 165,150 $$

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