Compute _n = 1^20 n + 3n.

We have that align* _n = 1^20 &= 41 52 63 74 2017 2118 2219 2320 &= 21 22 231 2 3 = 1771. align*

jmc

algebra intermediate

Problem

Compute

n = 1 \prod 20 \frac{n + 3}{n} .

Solution

We have that

n = 1 \prod 20 = \frac{4}{1} \cdot \frac{5}{2} \cdot \frac{6}{3} \cdot \frac{7}{4} \dots \frac{20}{17} \cdot \frac{21}{18} \cdot \frac{22}{19} \cdot \frac{23}{20} = \frac{21 \cdot 22 \cdot 23}{1 \cdot 2 \cdot 3} = 1771 .

Final answer

1771