Question 1 and 2, Exercise 4.8

Solutions of Question 1 and 2 of Exercise 4.8 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Using the method of difference, find the sum of the series: 3+7+13+21+ to n term.

Solution.

Let Sn=3+7+13+21+31++Tn Also Sn=3+7+13+21++Tn1+Tn. Subtracting second expression from the first expression, we have SnSn=3+7+13+21+31++Tn(3+7+13+21++Tn1+Tn) 0=3+(73)+(137)+(2113)+(3121)++(TnTn1)Tn.0=3+(4+6+8+10+ up to (n1) terms )Tn Then Tn=3+(4+6+8+10+ up to (n1) terms )=3+n12[2(4)+(n11)(2)](Sn=n2[2a+(n1)d])=3+n12[8+(n2)(2)]=3+n12[2n+4]=3+(n1)(n+2)=3+n2n+2n2=n2+n+1 Thus, the kth term of series: Tk=k2++k+1 Now taking sum, we get Sn=nk=1Tk=nk=1(k2++k+1)=nk=1k2+nk=1k+nk=11=n(n+1)(2n+1)6+n(n+1)2+n=n6[(n+1)(2n+1)+3(n+1)+6]=n6(2n2+2n+n+1+3n+3+6)=n6(2n2+6n+10)=n3(n2+3n+5) Hence the sum of given series is n3(n2+3n+5). GOOD m(

Using the method of difference, find the sum of the series: 1+4+10+22+ to n term.

Solution.

Let Sn=1+4+10+22+46++Tn Also Sn=1+4+10+22++Tn1+Tn. Subtracting second expression from the first expression, we have SnSn=1+4+10+22+46++Tn(1+4+10+22++Tn1+Tn) 0=1+(41)+(104)+(2210)+(4633)++(TnTn1)Tn.0=1+(3+6+12+24+ up to (n1) terms )Tn Then Tn=1+(3+6+12+24+ up to (n1) terms )=1+3(2n11)21(Sn=a(rn1)r1)=1+3(2n11)=1+32n13=32n12. Thus, the kth term of series: Tk=32k12 Now taking sum, we get Sn=nk=1Tk=nk=1(32k12)=3nk=12k12nk=11=3(1+2+22+23+... up to n terms)2n=3(1)(2n1)212n=3(2n1)2n Hence the sum of given series is 3(2n1)2n. GOOD m(