Sequence summation GRE question 
This problem is giving me a lot of trouble. the only way i can think of doing it is to take the integeral a(k) from 1 to 100 but that is definitely not what i am supposed to do since its for the GRE and no knowledge of calculus is necessary. . . helpppp

If you write out the first few terms of the series, you'll see that $$a_1+a_2+a_3+\cdots+a_{99}+a_{100}$$$$=\left(1-\frac{1}{2}\right)+ \left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3} - \frac{1}{4} \right) \cdots+\left(\frac{1}{99} - \frac{1}{100}\right) + \left( \frac{1}{100} - \frac{1}{101} \right).$$
Rearranging the parentheses yields $$1 + \left(-\frac{1}{2} + \frac{1}{2}\right) + \left(-\frac{1}{3} + \frac{1}{3}\right) + \cdots +\left(-\frac{1}{100} + \frac{1}{100} \right) - \frac{1}{101}$$ $$=1-\frac{1}{101} = \frac{100}{101}.$$

You can confirm that the general formula for the $n$th sum is $$\sum_{k=1}^{n}\frac{1}{k} - \frac{1}{k+1}=1-\frac{1}{n+1}$$ by induction, since it is clearly true for $n=1$, and then assuming that it holds for $n$, we can see that $$\sum_{k=1}^{n+1}\frac{1}{k} - \frac{1}{k+1}=\sum_{k=1}^{n}\left(\frac{1}{k} - \frac{1}{k+1}\right) + \left(\frac{1}{n+1} - \frac{1}{n+2} \right)$$
so from our inductive hypothesis we get
$$\left(1-\frac{1}{n+1}\right)+\left(\frac{1}{n+1} - \frac{1}{n+2}\right)$$
$$=1-\frac{1}{n+2}.$$
