Question

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

Answer 1

`"see explanation"`

Explanation:

`"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)`

Answer 2

Refer to the Proof given in the Explanation.

Explanation:

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.!