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

Responda:

#"see explanation"#

Explicação:

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

#"this involves the following steps "#

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

#‚ÄĘ " assume the result is true for n = k"#

#‚ÄĘ " 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)#