Answer:
\(\sum _{i=2}^73i-15=-9\)
Step-by-step explanation:
Given : \(\sum _{i=2}^73i-15\)
We have to evaluate the given summation.
Consider \(\sum _{i=2}^73i-15\)
using, we have,
\(\sum _{k=m}^n\:=\:\sum _{k=1}^n\:-\:\sum _{k=1}^{m-1}\)
\(=\sum _{i=1}^73i-15-\sum _{i=1}^13i-15\) ..............(1)
Consider, \(\sum _{i=1}^73i-15\)
Apply sum rule,
\(\quad \sum a_n+b_n=\sum a_n+\sum b_n\) , we have,
\(=\sum _{i=1}^73i-\sum _{i=1}^715\)
Now, first consider, \(\sum _{i=1}^73i\)
Apply constant multiplication rule,\(\quad \sum c\cdot a_n=c\cdot \sum a_n\) we have,
\(=3\cdot \sum \:_{i=1}^7i\)
\(=3\cdot \:28\\\\=84\)
Also, \(\sum _{i=1}^715=105\)
Thus, \(\sum _{i=1}^73i-15=84-105=-21\)
Similarly, \(\sum _{i=1}^13i-15=-12\)
Substitute in (1) , we have,
Thus, \(\sum _{i=2}^73i-15=-21-(-12)=-21+12=-9\)
Thus, \(\sum _{i=2}^73i-15=-9\)
