f(l)=0 f(n+1)−f(n)=5n f(2)−f(1)=5⇒f(2)=5 f(3)−f(2)=10⇒f(3)=15 f(4)−f(3)=15⇒f(4)=30 f(5)−f(4)=20⇒f(5)=50 So, x0,5,15,30,50 ⏟51 difference again 555←101520 Sn=T1+T2+T3+...+Tn Sn=T1+T2+...+Tn−1+Tn On subtracting, we get 0=T1+(T2−T1)+(T3−T2)+...+(Tn−Tn−1)−Tn Tn=T1+5+10+...+5(n−1) Tn=0+5[1+2+...+(n−1)] Tn=