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Solution :

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