Can you prove that #1^2+2^2+3^2+...+n^2=1/6n(n+1)(2n+1)#?
Risposta:
#"see explanation"#
Spiegazione:
#"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)#