In ascending order, each positive number [tex]n[/tex] appears [tex]n[/tex] times in a sequence [tex]a_{n}[/tex].
[tex]\text{$1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \cdots.$}[/tex]
Find the following value.
[tex]\Huge\text{$\rightarrow\displaystyle\sum^{7}_{k=1}a_{10^{k}}.$}[/tex]
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[tex]\Large\text{\underline{\underline{Explanation}}}[/tex]
[tex]\Large\text{$\underline{\underline{\text{Step 1.}}}$}[/tex]
Since a natural number [tex]n[/tex] appears [tex]n[/tex] times, each [tex]n[/tex] will stack up and result in -
[tex]\text{$\cdots\longrightarrow\boxed{\displaystyle\left(\sum^{n-1}_{k=1}k\right)+1\leq\text{(Term number)}\leq\displaystyle\sum^{n}_{k=1}k.}$}[/tex]
Let's simplify this inequality.
[tex]\text{$\cdots\longrightarrow\boxed{\dfrac{n(n-1)}{2}+1\leq\text{(Term number)}\leq\displaystyle\dfrac{n(n+1)}{2}.}$}[/tex]
Now, we need to know what [tex]\Large\text{$a_{10^{k}}$}[/tex] is.
Let's use the previous inequality we found.
[tex]\Large\text{$\underline{\underline{\text{Step 2.}}}$}[/tex]
From the previous inequality, -
[tex]\text{$\cdots\longrightarrow\boxed{\dfrac{n(n-1)}{2}+1\leq10^{k}\leq\displaystyle\dfrac{n(n+1)}{2}}$}[/tex]
[tex]\text{$\cdots\longrightarrow\begin{cases} & n^{2}-n-2\cdot10^{k}+1\leq0 \\ & n^{2}+n-2\cdot10^{k}\geq0 \end{cases}$}[/tex]
[tex]\text{$\cdots\longrightarrow\begin{cases}&n\leq\dfrac{1+\sqrt{8\cdot10^{k}-3}}{2} \\\\ & n\geq\dfrac{-1+\sqrt{8\cdot10^{k}+1}}{2} \end{cases}$}[/tex]
[tex]\text{$\cdots\longrightarrow\boxed{\dfrac{-1+\sqrt{8\cdot10^{k}+1}}{2}\leq n\leq\dfrac{1+\sqrt{8\cdot10^{k}-3}}{2}.}$}[/tex]
[tex]\Large\text{$\underline{\underline{\text{Step 3.}}}$}[/tex]
So, -
[tex]\text{$k=1\implies n=4.$}[/tex]
[tex]\text{$k=2\implies n=14.$}[/tex]
[tex]\text{$k=3\implies n=32.$}[/tex]
[tex]\text{$k=4\implies n=100.$}[/tex]
[tex]\text{$k=5\implies n=316.$}[/tex]
[tex]\text{$k=6\implies n=1000.$}[/tex]
[tex]\text{$k=7\implies n=3162.$}[/tex]
[tex]\Large\text{$\underline{\underline{\text{Final answer}}}$}[/tex]
Hence, -
[tex]\Large\text{$\cdots\longrightarrow\boxed{\displaystyle\sum^{7}_{k=1}a_{10^{k}}=4628.}$}[/tex]