Question

Can someone help me walk through the steps of this problem? to help with future ones?

Answer 1

Evaluate \[\sum_{n=1}^{4} 3n+1\]
a) 13
b) 34
c) 3640
d) 1/13

Answer 2

It's A

Answer 3

I lied it's 34.. B

Answer 4

\[\sum_{n=1}^{4} 3n+1=\big[3(1)+1\big]+\big[3(2)+1\big]+\big[3(3)+1\big]+\big[3(4)+1\big]\\\qquad\quad\qquad=\]

Answer 5

a simpler example

\[\sum_{n=4}^6n+1 =[4+1]+[5+1]+[6+1]=5+6+7=18\]

Answer 6

oh, okay thank you for showing me the steps to it now it makes much more sense to how to look and solve a problem like this. Thank you

Answer 7

so 34, thank you

Answer 8

yes!

Answer 9

The general sum

\[\sum_{n=a}^Nf(n) = f(a)+f(a+1)+f(a+2)+\cdots+f(N-1)+f(N)\]