How do you evaluate #sum_(n=3)^8 ( n+25)#?

1 Answer
Jun 13, 2016

# 183.#

Explanation:

Reqd. Sum #= sum_(n=3)^(n=8)(n+25)=sum_(n=3)^(n=8)n+sum_(n=3)^(n=8)25#
#={sum_(n=1)^(n=8)n-sum_(n=1)^(n=2)n}+{25+25+...+25," 6 terms}"#
#={(n*(n+1))/2}_(n=8)-{(n*(n+1))/2}_(n=2)+25*6#
#=(8*9)/2-(2*3)/2+150#
#=36-3+150#
#=183.#