Question

Find the sum of first n terms of the series 3+7+13+21+31..

Answer 1

Given series is is written asSn=3+7+13+21+31+.................+tn+0Sn=3+7+13+21+31+.................+tn+0..........(i)Sn=0+3+7+13+21+31+.................+tnSn=0+3+7+13+21+31+.................+tn.........(ii)Subtracting (i) - (ii) we get0=3+(4+6+8+10+............(n−1)terms)−tn0=3+(4+6+8+10+............(n−1)terms)−tn⇒tn=3+(4+6+8+10+.........(n−1)terms)⇒tn=3+(4+6+8+10+.........(n−1)terms)4+6+8+..........(n−1)terms4+6+8+..........(n−1)terms is an A.P.with 1st1st term =a=4=a=4 common difference =d=2=d=2 and no. of terms =n−1=n−1We know that sum of n−1n−1 terms of an A.P. =n−12=n−12[2a+(n−2)d][2a+(n−2)d]∴4+6+8+......=n−12∴4+6+8+......=n−12[2×4+(n−1−1)2][2×4+(n−1−1)2]=n−12=n−12[8+(n−2)2]=(n−1)(2+n)=n2+n−2[8+(n−2)2]=(n−1)(2+n)=n2+n−2Substituting this value in tntn we get⇒tn=3+(n2+n−2)⇒tn=3+(n2+n−2)i.e.,i.e., tn=n2+n+1tn=n2+n+1Step 2Now for any series Sum of nn terms =Sn=∑tn=Sn=∑tn⇒Sn=∑(n2+n+1)⇒Sn=∑(n2+n+1)=∑n2+∑n+∑1=∑n2+∑n+∑1=n(n+1)(2n+1)6+n(n+1)2+=n(n+1)(2n+1)6+n(n+1)2+nn=n[(n+1)(2n+1)6+n+12+=n[(n+1)(2n+1)6+n+12+1]1]⇒Sn=n6⇒Sn=n6[2n2+3n+1+3n+3+6][2n2+3n+1+3n+3+6].⇒Sn=n6⇒Sn=n6[2n2+6n+10]=n3[2n2+6n+10]=n3[n2+3n+5]