In the sequence $a_{1}, a_{2}, a_{3}, ..., a_{100}$, the $k$th term is defined by $$a_{k} = \frac{1}{k} - \frac{1}{k+1}$$ for all integers $k$ from $1$ through $100$. What is the sum of $100$ terms of this sequence?
The answer given is $\frac{100}{101}$, but I am not sure how.
So far I am have plugged in the values of $k's$ and have the following values $$\frac{1}{2}, \frac{1}{6}, \frac{1}{12}, \frac{1}{12}, \frac{1}{20}, \frac{1}{30},....$$
The numerator makes sense to me as it's just $1$ and $100 \times 1 = 100$, however, I am not sure about the denominator.
$\endgroup$ 43 Answers
$\begingroup$$$a_1+a_2+a_3+...+a_{100}=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{99}-\frac{1}{100}\right)+\left(\frac{1}{100}-\frac{1}{101}\right)=1-\frac{1}{101}=\frac{100}{101}$$
Note that $-\frac{1}{2}$ cancel $\frac{1}{2}$, $-\frac{1}{3}$ cancel $\frac{1}{3}$ and go on until $-\frac{1}{100}$ cancel $\frac{1}{100}$.
$\endgroup$ $\begingroup$$$a_{100} = \left( \frac11 - \frac12 \right)+\left( \frac12 - \frac13 \right)+ \left( \frac13 - \frac14 \right)+\cdots + \left( \frac1{100} - \frac1{101} \right) \\= \frac11 + \left( -\frac12+\frac12\right)+\left( -\frac13+\frac13\right) +\cdots +\left( -\frac1{100}+\frac1{100}\right) -\frac1{101} = 1 - \frac1{101} $$
$\endgroup$ $\begingroup$Such a sum is called telescop sum. We have $$\sum_{k=1}^{100} \left( \frac{1}{k}- \frac{1}{k+1} \right) \stackrel{\star}{=}1 - \frac{1}{101}.$$ In $(\star)$ we split the sum into two and make an index shift.
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