IF `f(r )=1+(1)/(2)+(1)/(3)+"...."+(1)/(r )` and `f(0)=0`, find `sum _(r=1)^(n)(2r+1)f(r )`.

Explanation

Since, `sum _(r=1)^(n)(2r+1)f(r )`
`=sum _(r=1)^(n)(r^(2)+2r+1-r^2))f(r )=sum _(r=1)^(n){(r+1)^(2)-r^(2)}f(r )`
`=sum _(r=1)^(n){(r+1)^(2)f(r )-(r+1)^(2)f(r+1)+(r+1)^(2)f(r+1)-r^(2)f(r )}`
`=sum _(r=1)^(n)(r+1)^(2){f(r )-(r+1)}+sum _(r=1)^(n){(r+1)^(2)f(r+1)-r^(2)f(r )}`
`=-sum _(r=1)^(n)((r+1)^(2))/((r+1))+sum _(r=1)^(n)(r+1)^(2)f(r+1)+(n+1)^(2) f(n+1)-sum_r^(r=1)r^(2)f(r)[:.f(r+1)-f(r)=(1)/(r+1)]`
`=-sum _(r=1)^(n)(r+1)+{2^(2)f(2)+3^(2)f(3)+"...."n^(2)f(n)}+(n+1)^(2)f(n+1)={1^(2)f(1)+2^(2)f(2)+3^(2)+"...."n^(2)f(n)}`
`=-sum _(r=1)^(n)r-sum _(r=1)^(n)1+(n+1)^(2)f(n+1)-1^(2)f(1)`
`=(n(n+1))/(2)-n+(n+1)^(2)f(n+1)-f(1)`
`=(n+1)^(2)f(n+1)-(n(n+1))/(2)-1 " "[:.f(1)=1]`
`=(n+1)^(2)f(n+1)-((n^(2)+3n+2))/(2)`
Hence, this is the required result.