Use sigma notation to write the sum. $$\frac { 1 } { 3 \cdot 2 } + \frac { 1 } { 4 \cdot 3 } + \ldots + \frac { 1 } { 8 \cdot 7 }$$
A) $$\sum _ { n = 1 } ^ { 6 } \frac { 1 } { ( n + 2 ) ! }$$
B) $\sum _ { n = 0 } ^ { 5 } \frac { 1 } { ( n + 1 ) ( n + 2 ) }$
C) $$\sum _ { n = 0 } ^ { 6 } \frac { 1 } { n ( n + 1 ) }$$
D) $\sum _ { n = 1 } ^ { 4 } \frac { 1 } { ( n + 1 ) ( n + 2 ) }$
E) $\sum _ { n = 1 } ^ { 6 } \frac { 1 } { ( n + 1 ) ( n + 2 ) }$

Question

Use sigma notation to write the sum. $$\frac { 1 } { 3 \cdot 2 } + \frac { 1 } { 4 \cdot 3 } + \ldots + \frac { 1 } { 8 \cdot 7 }$$
A) $$\sum _ { n = 1 } ^ { 6 } \frac { 1 } { ( n + 2 ) ! }$$
B) $\sum _ { n = 0 } ^ { 5 } \frac { 1 } { ( n + 1 ) ( n + 2 ) }$ ​
C) $$\sum _ { n = 0 } ^ { 6 } \frac { 1 } { n ( n + 1 ) }$$ ​
D) $\sum _ { n = 1 } ^ { 4 } \frac { 1 } { ( n + 1 ) ( n + 2 ) }$ ​
E) $\sum _ { n = 1 } ^ { 6 } \frac { 1 } { ( n + 1 ) ( n + 2 ) }$ ​

Quizplus · Accepted Answer

The given sum can be represented as $\sum _ { n = 1 } ^ { 6 } \frac { 1 } { ( n + 1 ) ( n + 2 ) }$ because it starts with $n = 1$ (which corresponds to the 3 in the denominator of the first term) and ends with $n = 6$ (which corresponds to the 8 in the denominator of the last term), and each term is of the form $\frac { 1 } { ( n + 1 ) ( n + 2 ) }$.

Use Sigma Notation to Write the Sum 13⋅2+14⋅3+…+18⋅7\frac { 1 } { 3 \cdot 2 } + \frac { 1 } { 4 \cdot 3 } + \ldots + \frac { 1 } { 8 \cdot 7 }3⋅21​+4⋅31​+…+8⋅71​

Use Sigma Notation to Write the Sum $\frac { 1 } { 3 \cdot 2 } + \frac { 1 } { 4 \cdot 3 } + \ldots + \frac { 1 } { 8 \cdot 7 }$