Sigma Notation help?

Question

To set this out, E = the sigma sign, the number before it is the number usally above it and the number after it will be the number usually underneath and the number in brackets can be the number next to the E.

Is "6En=1 (n²) a arithmeic progression or geometric progression?

I know the sum = 91, the sum answer if I use the arithmetic forumula =111 so its not that but if I try the geometric formula i don't know the common ratio?

Thankyou..helps heaps

Answer 1

It is neither an arithmetic nor a geometric progression. An arithmetic progression has a common difference between terms, and a geometric progression has a common ratio. Neither applies when you are summing exponential terms.

There is in fact a formula for summing squares up to M. Using your notation:
MEn=1(n^2) = M*(M+1)*(2M+1)/6

If you plug in M=6 as in your example, you will see that it gives you the correct value of 91.

Answer 2

Hi,

It is neither geometric nor arithmetic. If you let n = the integers from 1 through 6, the values of n^2 are 1, 4, 9, 16, 25, and 36. You can see these do not have a common difference that you can add every time. You can also see that you can't multiply by the same common ratio to go from one term to the next. Therefore, it's not arithmetic or geometric. It's simply the sum of perfect squares.

I hope this helps!!

Answer 3

Neither. The progression is 1, 4, 9, 16, 25, 36. It's not arithmetic because there is no similar added number (4-1=3, 9-4=5) and it is not geometric because there is no common ratio (4/1=4, 9/4=2.25). It's closest to a p-series, but I only know how to do that stuff when the series goes to infinity. If the series is finite your best bet is to just add it up by hand.

Answer 4

It is neither a geometric nor an arithmetic sequence.

The sum of the first n squares (1^1 + 2^2 + ... + n^2) can be calculated with this equation: nE1 (n^2) = (n)(n+1)(2n+1)/6.

For instance, the first 6 squares sum to (6)(6+1)(2*6+1)/6 = 91.

To prove the above formula is true requires induction on n after showing that it holds for the sum of the first 1 square(s).

Answer 5

a_1 = -2 = (-1)^1 * 2 = (-1)^1 * 2^1 a_2 = 4 = (-1)^2 * 4 = (-1)^2 * 2^2 a_3 = -8 = (-1)^3 * 8 = (-1)^3 * 2^3 a_4 = 16 = (-1)^4 * 16 = (-1)^4 * 2^4 a_5 = -32 = (-1)^5 * 32 = (-1)^5 * 2^5 a_6 = 64 = (-1)^6 * 64 = (-1)^6 * 2^6 Let k = 1, 2, 3, 4, 5, or 6 The general formula looks like a_k = (-1)^k * 2^k So we can apply the sigma notation as k ranges from 1 to 6: SIGMA[(-1)^k * 2^k] from k=1 to k=6

Answer 6

They fact that it is a power indicates it is geometric.