(i) `1^(2) +2^(2) +3^(2) +…+n^(2)`
Consider the identity `.(k+1)^(3)-K^(2)=3k^(2)+3k+1`
On putting k= 1,2,3…(n-1) ,n uccessively ,we get
`2^(3)-1^(3) =3.1^(2) +3.1+1`
`3^(3) -2^(3)=3.2^(2)+3.2+1`
`4^(3) -3^(3) +3.3+1`
`{:(...,...,...),(...,...,...):}`
`n^(3)-(n-1)^(3)=3.(n-1)^(2) +3.(n-1)+1`
` (n+1)^(3) -n^(3) =3.n+1`
Adding Columnwise .we get
`n^(3) +3n^(2) +3n=3(sum_(r=1)^(n) r^(2))+3(n(n+1))/(2) +n [ :' sum _(r-1)^(n) r^(2) =(n(n+1))/(2) ]`
` implies 3(sum_(r=1)^(n) r^(2))=n^(3) +3n^(2) 3n-(3n(n+1))/(2)+n`
`implies (sum _(r=1) ^(n) r^(2) )=(2n^(3)+3n^(2) +n)/(2) =(n(n+1)(2n+1))/(2)`
` implies sum _(r=1)^(n)r^(2) =(n(n+1))(2n+1))/(6) `
hence ` sum _(r=1)^(n)r^(2) =1^(2)+2^(2) +...+n^(2) =(n(n+1)(2n+1))/(6)`
(ii) `1^(3)+2^(3) +...+ n^(3)`
Consider the identity `(K+1)^(4)-k^(4)=4k^(3)+6k^(2)+4k+1 `
on putting `k=1,2,3,...(n-1)` n successively we get
`2^(4)-1^(4)=4.1^(3)+6.1^(2)+4.1+1`
`3^(4)-2^(4)=4.2^(3)+6.2^(2) +4.2+1`
`4^(4)-3^(4)=4.3^(3)+6.3^(2)+4.3+1`
`{:(...,...,...),(...,...,...):}`
`n^(4)-(n-1)^(4) +6(n-1)^(2) +4(n-1)+1`
`(n+1)^(4) -n^(4) =4.n^(3) +6.n^(2) +4.n+1`
Adding colurwise , we get
`(n+1)^(4) -1^(4) .(1^(3) +2^(3) +.....+n^(3) )+6(1^(2) +2^(2) +3^(3) +....+n^(2)) +4(1+2+3+....+n)+(1+1+...+1)` n terms
`implies n^(4) +4n^(3) +6n^(2) =4n=4(sum _(r=1)^(n) r^(3))+6(sum _(r=1)^(n) r^(2) ) +4(sum _(r=1)^nr) +n`
`implies n^(4)+4n^(3) +6n^(2) +4n=4(sum+(r=1)^(n)r^(3)) +6[(n(n+1)(2n+1))/(6)]+4[(n(n+1))/(2)]+n`
`implies sum _(r=1)^(n)r^(3)=(n^(2)(n+1)^(2)]/(4)`
`implies sum _(r=10)^(n) r^(3) =[(n(n+1))/(2)]^(2) =(sum _(r=1)^(n) r)^(2)`
hence ` sum _(r=1) ^(n) r^(3)=1^(3) +2^(3) +...+n^(3) =[(n(n+1))/(2)]^(2)=(sum_(r=1)^(n) r)^(2)`
(iii) `2+4+6+...+2n=2[1+2+3+...+n]`
`=2xx(n(n+1))/(2)=n+1)`
`s_(n) =1+2+3+...+n`
(iv) let `S_(n) 1+2+3+...+n`
Clearly it is an arithmetc series with first term ,a=1,
common difference , `d=1`
and last term = n
`S_(n) =(n)/(2) (1+n)=(n(n+1))/(2)`
hence `1+2+3+...+n=(n(n+1))/(2)`
