Write the sum using sigma notation. $\frac { 1 } { 1 \cdot 2 } + \frac { 1 } { 2 \cdot 3 } + \frac { 1 } { 3 \cdot 4 } + \ldots + \frac { 1 } { 999 \cdot 1000 }$
A) $\sum _ { n = 1 } ^ { 1000 } \frac { 1 } { n ( n + 1 ) }$

B) $\sum _ { n = 1 } ^ { 1000 } \frac { 1 } { n ( n - 1 ) }$

C) $\sum _ { n = 1 } ^ { 999 } \frac { 1 } { n ( n + 1 ) }$

D) $\sum _ { n = 1 } ^ { 1001 } \frac { 1 } { n ( n + 1 ) }$

E) $\sum _ { n = 1 } ^ { 999 } \frac { 1 } { n ( n - 1 ) }$

Question

Write the sum using sigma notation. $\frac { 1 } { 1 \cdot 2 } + \frac { 1 } { 2 \cdot 3 } + \frac { 1 } { 3 \cdot 4 } + \ldots + \frac { 1 } { 999 \cdot 1000 }$
A) $\sum _ { n = 1 } ^ { 1000 } \frac { 1 } { n ( n + 1 ) }$

B) $\sum _ { n = 1 } ^ { 1000 } \frac { 1 } { n ( n - 1 ) }$

C) $\sum _ { n = 1 } ^ { 999 } \frac { 1 } { n ( n + 1 ) }$

D) $\sum _ { n = 1 } ^ { 1001 } \frac { 1 } { n ( n + 1 ) }$

E) $\sum _ { n = 1 } ^ { 999 } \frac { 1 } { n ( n - 1 ) }$

Quizplus · Accepted Answer

We start by observing that in each fraction, the denominator is of the form $n(n+1)$. We can try to split this into partial fractions, to see if the sum telescopes. In other words, we want to find constants $a_n$ and $b_n$ such that $$\frac {1} {n ( n + 1 ) }=\frac {a_n} {n}+\frac {b_n} {n+1}.$$Solving for $a_n$ and $b_n$ by clearing denominators gives us $a_n=\frac{1}{n}-\frac{1}{n+1}$, and $b_n=\frac{1}{n+1}-\frac{1}{n+2}$. Hence, the sum rewrites as $$(\frac 11 - \frac 12) + (\frac 12 - \frac 23) + (\frac 13 - \frac 24) + \cdots + (\frac 1{999} - \frac{1}{1000}).$$All of the other answer choices are incorrect, because they do not have the alternating $\pm$ that we see while we are telescoping the initial expression.

Write the Sum Using Sigma Notation 11⋅2+12⋅3+13⋅4+…+1999⋅1000\frac { 1 } { 1 \cdot 2 } + \frac { 1 } { 2 \cdot 3 } + \frac { 1 } { 3 \cdot 4 } + \ldots + \frac { 1 } { 999 \cdot 1000 }1⋅21​+2⋅31​+3⋅41​+…+999⋅10001​

Write the Sum Using Sigma Notation $\frac { 1 } { 1 \cdot 2 } + \frac { 1 } { 2 \cdot 3 } + \frac { 1 } { 3 \cdot 4 } + \ldots + \frac { 1 } { 999 \cdot 1000 }$