475,106 views
40 votes
40 votes
Evaluate: sigma 32 and n=1 (2n+8)

User Squid
by
3.5k points

2 Answers

25 votes
25 votes

Answer:

Not sure but I think its 4n+90)

Explanation:

don't go with my answer to be safe

User Rishi Agarwal
by
3.1k points
20 votes
20 votes

Answer:


\sum_(n=1)^(32) 2n+8 =1312

Explanation:

I assume you are looking for
\sum_(n=1)^(32) 2n+8 \\:


\sum_(n=1)^(32) 2n+8 \\


\sum_(n=1)^(32) 2n+\sum_(n=1)^(32) 8 \\


2\sum_(n=1)^(32) n+32(8)


2((32(32+1))/(2))+256


2(528)+256


1056+256


1312

User Patrick Williams
by
3.2k points