Can you prove that #1^2+2^2+3^2+...+n^2=1/6n(n+1)(2n+1)#?

#"using the method of "color(blue)"proof by induction"#

#"this involves the following steps "#

#• " prove true for some value, say n = 1"#

#• " assume the result is true for n = k"#

#• " prove true for n = k + 1"#

#n=1toLHS=1^2=1#

#"and RHS " =1/6(1+1)(2+1)=1#

#rArrcolor(red)"result is true for n = 1"#

#"assume result is true for n = k"#

#color(magenta)"assume " 1^2+2^2+ .... +k^2=1/6k(k+1)(2k+1)#

#"prove true for n = k + 1"#

#1^2+2^2+..+k^2+(k+1)^2=1/6k(k+1)(2k+1)+(k+1)^2#

#=1/6(k+1)[k(2k+1)+6(k+1)]#

#=1/6(k+1)(2k^2+7k+6)#

#=1/6(k+1)(k+2)(2k+3)#

#=1/6n(n+1)(2n+1)to" with " n=k+1#

#rArrcolor(red)"result is true for n = k + 1"#

#rArr1^2+2^2+3^2+....+n^2=1/6n(n+1)(2n+1)#

Let, #S_n=1^2+2^2+3^2+...+n^2, &, , f(n)=n^3, n in NNuu{0}.#

#:. f(n)-f(n-1)=n^3-(n-1)^3.#

#because, a^3-b^3=(a-b)(a^2+ab+b^2), f(n)-f(n-1),#

#={n-(n-1)}{n^2+n(n-1)+(n-1)^2},#

#=(1)(n^2+n^2-n+n^2-2n+1),#

# rArr f(n)-f(n-1)=n^3-(n-1)^3=3n^2-3n+1;(n in NNuu{0}.#

#n=1 rArr 1^3-0^3=3(1)^2-3(1)+1;#
#n=2 rArr 2^3-1^3=3(2)^2-3(2)+1;#
#n=3 rArr 3^3-2^3=3(2)^2-3(2)+1;#
#vdots vdots vdots vdots vdots vdots vdots vdots vdots vdots#
#n=n rArr n^3-(n-1)^3=3(n)^2-3(n)+1;#

#"Adding, "n^3-0^3=3{1^2+2^2+3^2+...+n^2}-3{1+2+3+...+n}+n,#

# :. n^3=3S_n-3Sigman+n, or, #

# n^3=3S_n-3/2n(n+1)+n, i.e.,#

#2n^3=6S_n-3n(n+1)+2n=6S_n-3n^2-3n+2n,#

# :. 2n^3+3n^2+n=6S_n,#

# :. 6S_n=n(2n^2+3n+1)=n(n+1)(2n+1),#

# rArr S_n=n/6(n+1)(2n+1).#

Enjoy Maths.!