There is the sequence of numbers $1, a_2, a_3, ...$ such that satisfies $1 \cdot a_2 \cdot a_3 \cdot ... \cdot a_n = n^2$, for every integer $n> 2$. Determine the value of $a_3 + a_5$.
Problem
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Tags: algebra
jasperE3
14.09.2021 23:48
Let $a_1=a_2=1$ and $a_n=\frac{n^2}3$ for $n>2$. Then $a_3+a_5=\frac{34}3$. Let $a_1=a_2=\frac13$ and $a_n=n^2$ for $n>2$. Then $a_3+a_5=34$.
parmenides51
14.09.2021 23:59
@ above, I had a huge typo, read $1$ instead of $a_1$ in the product
in terms of $a_2$ looking at official solutions
jasperE3
15.09.2021 00:05
Let $a_1=1$. Then: $$a_{n+1}=\frac{\displaystyle\prod_{i=1}^{n+1}a_i}{\displaystyle\prod_{i=1}^na_i}=\frac{(n+1)^2}{n^2}$$for $n>2$, so $a_5=\frac{25}{16}$. Also, $1\cdot a_2\cdot a_3=9$, hence $a_3=\frac9{a_2}$. Then $a_3+a_5=\boxed{\frac9{a_2}+\frac{25}{16}}$.
OlympusHero
15.09.2021 04:50
Note that $a_{n+1} \cdot n^2 = (n+1)^2$, so $a_{n+1}=\frac{(n+1)^2}{n^2}$ for all $n>2$. This means $a_5=\frac{25}{16}$. We also have $a_2a_3=9$, so $a_3=\frac{9}{a_2}$. The answer is $\boxed{\frac{25}{16}+\frac{9}{a_2}}$.