Question

Help please

Answer 1

I can't get sigma x^2 for the set of 50 numbers

Answer 2

5500 + 9600

Answer 3

I have researched this a while ago. there is actually an algorithm that determines this value.\[\sum_{k=m}^{n}k^2=\frac{1}{3}((n+1)^3-m^3)-\frac{1}{2}((n+1)^2-m^2)+\frac{1}{6}((n+1)-m)\]
in your case, m=1 and n=50.
\[\sum_{k=1}^{50}k^2=\frac{1}{3}(51^3-1^3)-\frac{1}{2}(51^2-1^2)+\frac{1}{6}(51-1)\]
\[a=51^3-1^3=(125000+3*2500+3*50+1)-1=132650\]
\[b=51^2-1^2=(2500 + 2*50 + 1) -1=2600\]
\[c=51-1=50\]
\[\frac{1}{3}a-\frac{1}{2}b+\frac{1}{6}c=\frac{265300-7800+50}{6}=\frac{257550}{6}=42925\]

Answer 4

therefore the answer is 42925. message me for an algorithm for any polynomial