### Instruction: Express the value of the following as a common fraction:
$\left(1-\frac12\right)\cdot\left(1-\frac13\right)\cdot\left(1-\frac14\right) \dotsm \left(1-\frac1{n+1}\right) \dotsm \left(1-\frac1{100}\right)$.### Response: I notice that each term inside the parentheses has the form $1-\frac1k$, which can be rewritten as $\frac{k-1}{k}$.
So I can rewrite the whole expression as a product of fractions: $\frac12\cdot\frac23\cdot\frac34\cdot\dotsm\cdot\frac{n}{n+1}\cdot\dotsm\cdot\frac{99}{100}$.
Now I see that this product has a lot of cancellation going on: the denominator of the first fraction cancels with the numerator of the second, the denominator of the second fraction cancels with the numerator of the third, and so on.
In fact, the only factors that do not cancel are the numerator of the first fraction and the denominator of the last fraction.