(i) Write `sum_(r=1)^(n)(r^(2)+2)` in expanded form.
(ii) Write the series `(1)/(3)+(2)/(4)+(3)/(5)+(4)/(6)+"…"+(n)/(n+2)` in sigma form.

Explanation

(i) On putting `r=1,2,3,4,"....","n in"(r^(2)+2),`
we get `3,6,11,18,"…",` `(n^(2)+2)`
Hence,`sum_(r=1)^(n)(r^(2)+2)=3+6+11+18+"...."+(n^(2)+2)`
(ii) The rth terms of series `=(r)/(r+2).`
Hence, the give series can be written as
`(1)/(3)+(2)/(4)+(3)/(5)+(4)/(6)+"...."+(n)/(n+2)=sum_(r=1)^(n)((r)/(r+2))`