\(1+2+3+....+(n-1)+n=\dfrac{n(n+1)}{2}\)
\(p+(p+1)+....+(q-1)+q= \dfrac{(q+p)(q-p+1)}{2}\)
\(1+3+5+....+(2n-3)+(2n-1) =n^2\)
\(2+4+6+....+(2n-2)+2n=n(n+1)\)
\(1^2+2^2+3^2+...+ (n-1)^2 +n^2= \dfrac{n(n+1)(2n+1)}{6}\)
\(1^3+2^3+3^3+....+(n-1)^3+n^3=\dfrac{n^2(n+1)^2}{4}\)
\(1^2+3^2+5^2+....+(2n-3)^2+(2n-1)^2=\dfrac{n(4n^2-1)}{3}\)
\(1^3+3^3+5^3+....+(2n-3)^3+(2n-1)^3=n^2(2n^2-1)\)
\(1^4+2^4+3^4+....+(n-1)^4 +n^4=\dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}\)
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