# Sigma Notation problem

1. Oct 19, 2007

### Vipul

[a]1. The problem statement, all variables and given/known data[/b]

Evaluate:
12
$$\sum$$ (6n + 1)
n=5

2. The attempt at the solution
So, how do i go about doing this? I tried finding the first three numbers of the series to find the difference by substituting n = 5,6,7 and then use the Sum formula S = n/2[2a + (n-1)d]. But the answer turned out to be wrong. The correct answer to this is 416.

Last edited: Oct 19, 2007
2. Oct 19, 2007

### qspeechc

Remember that you are summing from n=5 to n=12, not n=0 to n=12. Does that make a difference? Can't really tell because tyou never showed your working.

3. Oct 19, 2007

### Vipul

So i substituted n = 5,6,7
so the sequences is as follows: 31,37,43.
So from the sequence a = 31 and d = 6
Substituting into the formula S = n/2[2a + (n-1)d]

S = 12/2 [2(31) + (12-1)6]
S = 6[62 + 66]
S = 6 X 128
S = 768

EDIT : Found the solution. the number of terms is 8, not 12. Because you actually count the number of terms from 5 to 12.

Last edited: Oct 19, 2007
4. Oct 19, 2007

### qspeechc

Correct, you should remember that if you are summing from a to b there are (b-a+1) terms. If you take (b-a) terms you will be one short because you have not included a as your first term. Examine this, think of the difference as being equal to the number of bracketed things (if that makes sense):

b-a = (a+1),(a+2),(a+3),...(b)
b-a+1= [(a+1),(a+2),(a+3),...(b)]+1
OR:
b-a+1= (a),(a+1),(a+2),(a+3),...,(b)

Do you see why it should be b-a+1 and not b-a ?

I am sorry if I am labouring the point too much :(

5. Oct 19, 2007

### Vipul

Yes, got the point. Thanks a lot :D