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First, looking at it as a telescoping sum, you will get $$\sum_{i=1}^n((1+i)^3-i^3)=(1+n)^3-1.$$ On the other hand, you also have $$\sum_{i=1}^n((1+i)^3-i^3)=\sum_{i=1}^n(3i^2+3i+1)=3\sum_{i=1}^ni^2+3\sum_{i=1}^ni+n.$$ Using these two expressions, and the fact that $\sum_{i=1}^ni=\frac{n(n+1)}{2}$, you can now solve for $\sum_{i=1}^ni^2$.
I am studying clustering with K-Means algorithm and I got stumbled in the "inertia", or "within cluster sum of squares" part. First I would appreciate if anyone could explain me...
According to my calculations. The mark scheme states an alternative approach when calculating all the values of the ANOVA table. Σxt = 15207, Σx2 t = 6682987. Σ x t = 15207, Σ x t 2 = 6682987. From here this gives SST = Σx2 t − (Σxt)2/36 = 259296.75 S S T = Σ x t 2 − (Σ x t) 2 / 36 = 259296.75, and SSE = Σx2 t − (Σxi)2/12 ...
The following is a very general technique. In particular, it's probably not the best way to solve this sort of problem in the beginning. But it is worth knowing this technique in the long run. We use the result that: n − 1 ∑ k = 0(k i) = (n i + 1) Write: (2k + 1)2 = 8(k 2) + 8(k 1) + (k 0) Then n − 1 ∑ k = 0(2k + 1)2 = n − 1 ∑ k = 0 ...
Then I searched on the internet on how to calculate the sum of squares easily and found the below equation ...
First we note that y is linear in w, as y(x, w + μw ′) = ∑ i (wi + μw ′ i)xi = ∑ i wixi + μ∑ i w ′ ixi = y(x, w) + μy(x, w ′) Now we have for w, h ∈ Rm + 1 that E(w + h) = 1 2 N ∑ n = 1 (y(xn, w) + y(xn, h) − tn)2 = E(w) + N ∑ n = 1 (y(xn, w) − tn)y(xn, h) + N ∑ n = 1y(xn, h)2 = E(w) + N ∑ n = 1 (y(xn, w) − tn ...
I am trying to figure out how to get a sum of squares from a standard deviation. The standard deviations I have are $11.04$, $9.91$ and $9.43$. How do I calculate the sum of squares associated with...
For example, for p = 10 p = 10, the exact value is ≈ 22.46827819 ≈ 22.46827819 while the above approximation gives ≈ 22.46827983 ≈ 22.46827983. By itself, the first term already gives 21.0819 21.0819; the sum of first and second term gives ≈ 22.6629 ≈ 22.6629. For p = 100 p = 100, the approximation leads to 12 12 exact significant ...
I was trying to calculate the expected value of Treatment sum of squares and I saw this ... E(SSTr) = (I − 1)σ2 + J(∑I (μ2 i −μ2)) = (I − 1)σ2 + J(∑I (μi − μ)) =. My question is is this correct and how to prove that (I tried to come from righ to left but it didnt work out) ∑I (μ2 i −μ2) =∑I (μi − μ)2. I : is number ...
Any line that goes through the plotted pairs of points minimizes the sum of absolute values. The slopes of those lines range from $-0.25$ to $+1.25$. However, only the line that goes through the midpoints of the pairs of points minimizes the sum of squares.