<s>[INST] Confusion in fraction notation $$a_n = n\dfrac{n^2 + 5}{4}$$
In the above fraction series, for $n=3$ I think the answer should be $26/4$, while the answer in the answer book is $21/2$ (or $42/4$). I think the difference stems from how we treat the first $n$. In my understanding, the first number is a complete part and should be added to fraction, while the book treats it as part of fraction itself, thus multiplying it with $n^2+5$. 
So, I just want to understand which convention is correct.
This is from problem 6 in exercise 9.1 on page 180 of the book Sequences and Series.
Here is the answer sheet from the book (answer 6, 3rd element):


*

*$3,8,15,24,35$

*$\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},\dfrac{4}{5},\dfrac{5}{6}$

*$2, 4, 8, 16 \text{ and } 32$

*$-\dfrac{1}{6},\dfrac{1}{6},\dfrac{1}{2},\dfrac{5}{6},\dfrac{7}{6}$

*$25,-125,625,-3125,15625$

*$\dfrac{3}{2},\dfrac{9}{2},\dfrac{21}{2},21,\dfrac{75}{2}$

*$65, 93$

*$\dfrac{49}{128}$

*$729$

*$\dfrac{360}{23}$

*$3, 11, 35, 107, 323$; $3+11+35+107+323+...$

*$-1,\dfrac{-1}{2},\dfrac{-1}{6},\dfrac{-1}{24},\dfrac{-1}{120}$; $-1+(\dfrac{-1}{2})+(\dfrac{-1}{6})+(\dfrac{-1}{24})+(d\frac{-1}{120})+...$

*$2, 2, 1, 0, -1$; $2+2+1+0+(-1)+...$

*$1,2,\dfrac{3}{5},\dfrac{8}{5}$

 [/INST] $3 \times \frac{3^2+5}{4} = 3 \times \frac{9+5}{4} = 3\times \frac{14}{4} = 3\times \frac{7}{2} = \frac{21}{2}$
Normally (even though in calculator this is often not true) the convention is that a number on the side of a fraction is multiplying that fraction. There should be a "times" either a cross or a dot, however often you can omit it as a shortcut (in veritas, it is almost always omitted apart from very specific cases when someone wants to emphasise the steps as in my answer above)
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