Thread: Anyone intrested in checking sigma notation?

Could someone please check my work?

107
Σ (4+3i)
i=29

107 107
Σ (4+3i)=4*107+3Σi
i=29 29

=9737

Many thanks

This is what MathCad makes of your sum.

Originally Posted by stewpot

Code:

107
 Σ (4+3i)
i=29

107                  107
 Σ (4+3i) = 4*107 + 3 Σi
i=29                 29

I think you may have forgotten about your lower limit...?

You are not summing from i = 1 to i = 107; you are summing from i = 29 to i = 107. As such, you should have only 107 - 29 + 1 = 81 terms. This makes the first summation that you broke out, the "four times the summation", equal to 4*81 = 324, not 4*107 = 428.

The second summation can be represented as three times the difference of the sum of the first 107 numbers and the sum of the first 28 numbers.

Apply the formula, do the subtraction, and multiply by 3. Then add in the "four times" part.

Originally Posted by stewpot

Could someone please check my work?

107
Σ (4+3i)
i=29

107 107
Σ (4+3i)=4*107+3Σi
i=29 29

=9737

Many thanks

First, it would help if you figured out how many terms there will be. In this case there are 79.



.

This sum is an arithmetic series - so you solve it using the formula.







.

Brill, many thanks