Последовательность задана формулой
\(\displaystyle a_{n}=3n-1{\small.} \)
Выпишите первые \(\displaystyle 5\) членов последовательности.
\(\displaystyle a_1=\)\(\displaystyle {\small,}\) \(\displaystyle a_2=\)\(\displaystyle {\small,}\) \(\displaystyle a_3=\)\(\displaystyle {\small,}\) \(\displaystyle a_4=\)\(\displaystyle {\small,}\) \(\displaystyle a_5=\)\(\displaystyle {\small.}\)
Для нахождения первых \(\displaystyle 5\) членов последовательности подставим в формулу \(\displaystyle n\)-го члена
\(\displaystyle a_{\color{red}{n}}=3\color{red}{n}-1\)
натуральные числа от \(\displaystyle 1\) до \(\displaystyle 5 {\small.}\)
Получим:
| при \(\displaystyle \color{red}{n}=\color{red}{1}\) | \(\displaystyle a_{\color{red}{1}}=3 \cdot \color{red}{1}-1=2{\small;}\) | |
| при \(\displaystyle \color{red}{n}=\color{red}{2}\) | \(\displaystyle a_{\color{red}{2}}=3 \cdot \color{red}{2}-1=5{\small;}\) | |
| при \(\displaystyle \color{red}{n}=\color{red}{3}\) | \(\displaystyle a_{\color{red}{3}}=3 \cdot \color{red}{3}-1=8{\small;}\) | |
| при \(\displaystyle \color{red}{n}=\color{red}{4}\) | \(\displaystyle a_{\color{red}{4}}=3 \cdot \color{red}{4}-1=11{\small;}\) | |
| при \(\displaystyle \color{red}{n}=\color{red}{5}\) | \(\displaystyle a_{\color{red}{5}}=3 \cdot \color{red}{5}-1=14{\small.}\) |
Ответ: \(\displaystyle a_1=2{\small,} \,\, a_2=5{\small,} \,\, a_3=8{\small,} \,\, a_4=11{\small,} \,\, a_5=14{\small.}\)