∑k^4=1+∑(k+1)^4-(x+1)^4=1+∑k^4+4∑k^3+6∑k^2+4∑k+∑1-(x+1)^4-->(x+1)^4=1+4∑k^3+6∑k^2*+4∑k+∑1=1+∑k^3+x(x+1)(2x+1)+2x(x+1)+xSimplify,x^4+2x^3+x^2=4∑k^3=4( x(x+1)/2 )^2*Sum of squares underlined can be derived similarly as well, examining ∑k^3
√(1³) = 1√(1³+2³) = 1+2√(1³+2³+3³) = 1+2+3√(1³+2³+3³+4³) = 1+2+3+4
22:12 - 23 May 2026