Question

I just need to see if I'm doing this correctly.. question in the comments.

Answer 1

Find the sum of the finite series.\[\sum_{n=1}^{10}(3n-4)\] So do I just plug in 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 in for n and add them all together?

Answer 2

@iGreen

Answer 3

that will work, but there are shorter ways to do it.

Answer 4

what's the shorter way?

Answer 5

because right now, I have -1+2+5+8+11+14+17+20+23+24

Answer 6

=123

Answer 7

@phi

Answer 8

@ganeshie8

Answer 9

oops, not 24, 26

Answer 10

so it equals 125

Answer 11

the short way is write the problem as
\[ \sum_{n=1}^{10}(3n-4) = 3\sum_{n=1}^{10}n- \sum_{n=1}^{10} 4\]

the last summation is the sum of ten 4's, which is 10*4= 40. make it negative because we will be subtracting this: -40

the first summation is the sum of the first 10 numbers
there is a formula to find the sum of the first n numbers (starting with 1)
it is n(n+1)/2
here n is 10, so the sum of the first 10 numbers is 10*11/2 = 5*11= 55
now multiply that by 3: 3*55= 165
total is 165-40= 125