We define the sequences $a_n =\frac{n (n + 1)}{2}$ and $b_n = a_1 + a_2 +… + a_n$. Prove that there is no integer $n$ such that $b_n = 2017$.
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Tags: algebra, Sequences, Sequence
jasperE3
23.10.2021 07:52
$b_n$ is strictly increasing and $b_{21}=1771$ while $b_{22}=2024$.
BackToSchool
25.10.2021 00:19
Note that:
\begin{align*}
a_1 = \frac{1 \times 2}{2} &= \frac {1}{6} (1\times 2 \times 3 - 0 \times 1 \times 2) \\
a_2 = \frac{2 \times 3}{2} &= \frac {1}{6} (2\times 3 \times 4 - 1\times 2 \times 3) \\
a_3 = \frac{3 \times 4}{2} &= \frac {1}{6} (3\times 4 \times 5 - 2\times 3 \times 4) \\
\\
a_n = \frac{n (n + 1)}{2} &= \frac {1}{6} [n (n+1) (n+2) - (n-1) n (n+1)]
\end{align*}$$b_n=\sum_{i=1}^n a_i = \sum_{i=1}^n \frac {i(i+1)}{2}=\frac {1}{6} n (n+1) (n+2)$$$$6b_n = (n+1)^3-(n+1)$$Observe the units digit of cube, we have: $$n \rightarrow n^3 \Rightarrow (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) \rightarrow (0, 1, 8, 7, 4, 5, 6, 3, 2, 9)$$The units digit of $(n+1)^3-(n+1)$ cannot be $2$, then units digit of $b_n$ cannot be $7$, and thus $b_n$ cannot be $2017$.
Actually, $b_n$ cannot be equal to any number with units digit of $7$.