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1, 3, 7, 12, 19, ...
1, 7, 19, ...
1=1; 1+7=8; 1+7+19=27; ...
1, 8, 27, ...
The final sequence is the cube numbers. To show why, let \(n\) be an integer and follow through the process.
Cross out every third number:
1, 2, 3, 4, 5, 6, ..., 3n, \(3n+1\), \(3n+2\), ...
1, 2, 4, 5, ..., \(3n+1\), \(3n+2\), ...
Find the cumulative sums:
$$1=1$$
$$1+2=1+2=3$$
$$1+2+4=1+2+3+4-3=7$$
$$1+2+4+5=1+2+3+4+5-3=12$$
$$...$$
$$1+2+4+5+...+(3n+1)=\sum_{i=1}^{3n+1}-\sum_{i=1}^{n}3i$$
$$=\frac{1}{2}(3n+1)(3n+2)-\frac{3}{2}n(n+1)$$
$$=3n^2+3n+1$$
$$1+2+4+5+...+(3n+2)=3n^2+3n+1+(3n+2)$$
$$=3n^2+6n+3$$
$$...$$
1, 3, 7, 12, ..., \(3n^2+3n+1\), \(3n^2+6n+3\), ...
Cross out every second number, starting with the second:
1, 3, 7, 12, ..., \(3n^2+3n+1\), 3n2+6n+3, ...
1, 7, ..., \(3n^2+3n+1\), ...
Find the cumulative sums. The \(m\)th sum is:
$$\sum_{n=0}^{m}3n^2+3n+1$$
$$=3\sum_{n=0}^{m}n^2+3\sum_{n=0}^{m}n+\sum_{n=0}^{m}1$$
$$=\frac{3}{6}m(m+1)(2m+1)+\frac{3}{2}m(m+1)+m+1$$
$$=\frac{1}{2}(m+1)(m(2m+1)+3m+2)$$
$$=\frac{1}{2}(m+1)(2m^2+m+3m+2)$$
$$=\frac{1}{2}(m+1)(2m^2+4m+2)$$
$$=(m+1)(m^2+2m+1)$$
$$=(m+1)(m+1)^2$$
$$=(m+1)^3$$
Hence the numbers obtained are the cube numbers.
Extension
What happens if you cross out every third number starting at the second? Or every fifth number starting at the fifth? Or every \(n\)th number starting at the \(m\)th?
