Summation formula giving me different answer than expected

[vaderboi]

So I have this summation in sigma notation:

n = 4, i = 0, and a = i²

Based on my calculations, the summation should be equal to 30:

0² + 1² + 2² + 3² + 4² = 0 + 1 + 4 + 9 + 16 = 30

But when I input the values in to the summation formula I get 32:

n(a₀ + a_n) / 2

4(0 + 16) / 2

64 / 2 = 32

Are any of my calculations incorrect? I am confused as to what I am doing incorrectly.

 

[Harry_the_cat]

So I have this summation in sigma notation:

n = 4, i = 0, and a = i²

Based on my calculations, the summation should be equal to 30:

0² + 1² + 2² + 3² + 4² = 0 + 1 + 4 + 9 + 16 = 30

But when I input the values in to the summation formula I get 32:

n(a₀ + a_n) / 2

4(0 + 16) / 2

64 / 2 = 32

Are any of my calculations incorrect? I am confused as to what I am doing incorrectly.

The summation formula you are attempting to use applies only to arithmetic progressions.

0, 1, 4, 9, 16 is NOT an arithmetic progression (the differences between successive terms in an arithmetic progression are the same; here the differences (1-0=1, 4-1=3, 9-4=5,16-9=7 ) are NOT the same)

30 is the correct answer.

 

[vaderboi]

Ah! That makes sense. I thought I was going crazy. Is there a different formula for adding together successive values that don't share a common difference?

 

[Dr.Peterson]

So I have this summation in sigma notation:

n = 4, i = 0, and a = i²

Is there a different formula for adding together successive values that don't share a common difference?

For some finite series, there is a formula; for others, there is not. In general, they are difficult to find, if possible at all.

In this case, you have (I assume) \(\displaystyle \sum_{i=0}^{4}i^2\). It happens that there is a formula for it, which you might find by searching, or could derive it if you had learned appropriate methods; but almost certainly you are expected just to write out the terms and add them. The exercise is about understanding the notation, not about general formulas.

 

[Denis]

If you ask George Google, he'll quickly give you
the formula for sum of 1st n squares:
[n(n + 1)(2n + 1)] / 6

As you've been told, the 1st term = 1^2

If you're summing where 1st term > 1 (like sum of squares from 5 to 9)
then simply use the formula twice:
SUM(1 to 9) - SUM(1 to 4)