Question

prove by math induction :

1^2 +2^2 +3^2 + ... + n^2 = [n(n+1)(2n+1)]/6

Answer 1

first step n=1

1^2 = [1(1+1)(2*1+1)]/6
1 = [1*2*3]/6
1 = 6/6
1 = 1 so this is true

rewrite it for n = k and will get

1^2 +2^2 + ...+k^2 = [k(k+1)(2k+1)]/6 suppose this is true

and rewrite it for k=k+1 and prove it that is true -

Answer 2

\[1^2 +2^2 +3^2 + ... + \color{red}{ n^2} = \frac{\color{Red}{ n}(\color{Red}{n}+1)(2\color{Red}{n}+1) }{ 6 }\]
\[1^2 +2^2 +3^2 + ... + \color{red}{ k^2} = \frac{\color{Red}{ k}(\color{Red}{k}+1)(2\color{Red}{k}+1) }{ 6 }\]


n=k+1
\[1^2 +2^2 +3^2 + ... + \color{red}{ k^2}+ \color{blue}{(k+1)^2} = \frac{\color{blue}{ (k+1)}[\color{Red}{(k+1)}+1][2\color{Red}{(k+1)}+1] }{ 6 }\]

Answer 3

\[ \color{green}{1^2 +2^2 +3^2 + ... + \color{red}{ k^2}} +\color{blue}{(k+1)^2} = \frac{\color{blue}{( k+1)}[\color{Red}{(k+1)}+1][2\color{Red}{(k+1)}+1] }{ 6 }\]

1^2+2^2+3^2+......+k^2 = \[ 1^2+2^2+3^2+......+k^2 = \frac{ k(k+1)(2k+1) }{ 6 }\]
thus....
\[ \color{orange}{ \frac{ k(k+1)(2k+1) }{ 6 }} +\color{blue}{(k+1)^2} = \frac{\color{blue}{( k+1)}[\color{Red}{(k+1)}+1][2\color{Red}{(k+1)}+1] }{ 6 }\]

Answer 4

adding both sides (k+1)^2
\[1^2+2^2+...+k^2+(k+1)^2=\frac{ k(k+1)(2k+1 }{ 6 }+(k+1)^2\]
\[=(k+1)\left[ \frac{ k(2k+1)+6(k+1) }{ 6 } \right]=(k+1)\left[ \frac{ 2k^2+k+6k+6 }{ 6 } \right]\]
\[=(k+1)\left[ \frac{ 2k^2+7k+6 }{ 6 } \right]=(k+1)\left[\frac{ 2k^2+4k+3k+6 }{ 6 } \right]\]
\[=(k+1)\left[ \frac{ 2k(k+2)+3(k+2) }{ 6 } \right]=\frac{ (k+1)(k+2)(2k+3) }{ 6 }\]
\[=\frac{ (k+1)\left\{ \left( k+1 \right)+1 \right\}\left\{ 2(k+1)+1 \right\} }{ 6 }\]

Answer 5

People often forget to use LHS and RHS which is important when you are comparing the given induction to the simulated induction (i.e. using the n = k result).

Should technically add '= RHS' for completion.

Answer 6

@ParthKohli your opinion please about this above wrote proof by math induction is acceptably ? thank you

Answer 7

Yes, it's completely correct.