How do you find the sum of the arithmetic sequence #Sigma (2n+1)# from n=1 to 7?
1 Answer
Jan 15, 2018
Explanation:
#•color(white)(x)sum_(r=1)^n1=n#
#•color(white)(x)sum_(r=1)^nkr=ksum_(r=1)^n r=1/2n(n+1)#
#rArrsum_(r=1)^n(2n+1)#
#=2sum_(n=1)^7n+sum_(n=1)^7 1#
#=2xx1/2n(n+1)+n#
#=n^2+2n#
#"substitute " n=7to49+14=63#
#rArrsum_(n=1)^7(2n+1)=63#
#color(blue)"OR"#
#"since n is reasonably small we can use direct substitution"#
#"into the sum"#
#sum_(n=1)^7(2n+1)=3+5+7+9+11+13+15=63#
