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)