How do you work out the product of this sequence? I got a question in my maths paper and I didn't know how to answer it. This was the question:

What is: $(1+\frac{1}{2}) (1+\frac{1}{3}) (1+\frac{1}{4}) $... All the way up to 98 factors
a) What is the product up to 98 terms
b) What is the product up to n terms
 A: The question presented here seems to be a finite product. This can be written in the following notation:
$$\prod_{n=1}^{N}(1+\frac{1}{n+1})$$
Where $N = $ the term you wish to calculate the series up to. In case $A$, N = 98

Let's work out the series up to:
N=1: $(1+\frac{1}{2}) = \frac{3}{2}$
N=2: $ \frac{3}{2} * (1+\frac{1}{3}) = 2$
N=3: $ 2 * (1+\frac{1}{4}) =  \frac{5}{2}$
We can see that each term increments by $ \frac{1}{2}$ And so our N-th term can be written as:
$Tn = 0.5n + 1$
Therefore $T98 = 0.5(98) + 1 = 50$
A: Start with
Joshua Lochner's
product:
$\prod_{n=1}^{N}(1+\frac{1}{n+1})
$.
Then manipulate it like this:
$\begin{array}\\
\prod_{n=1}^{N}(1+\frac{1}{n+1})
&=\prod_{n=1}^{N}(\frac{n+2}{n+1})\\
&=\frac{\prod_{n=1}^{N}(n+2)}{\prod_{n=1}^{N}(n+1)}\\
&=\frac{\prod_{n=2}^{N+1}(n+1)}{\prod_{n=1}^{N}(n+1)}\\
&=\frac{\left(\prod_{n=2}^{N}(n+1)\right)(N+2)}{2\prod_{n=2}^{N}(n+1)}
\qquad\text{Isolate the common terms in each product}\\
&=\frac{(N+2)}{2}\\
&=\frac{N}{2}+1\\
\end{array}
$
