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.
Question 1
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+…+Tn−1+Tn.
Subtracting second expression from the first expression, we have
Sn−Sn=3+7+13+21+31+…+Tn−(3+7+13+21+…+Tn−1+Tn)
⟹0=3+(7−3)+(13−7)+(21−13)+(31−21)+…+(Tn−Tn−1)−Tn.⟹0=3+(4+6+8+10+… up to (n−1) terms )−Tn
Then
Tn=3+(4+6+8+10+… up to (n−1) terms )=3+n−12[2(4)+(n−1−1)(2)](∵Sn=n2[2a+(n−1)d])=3+n−12[8+(n−2)(2)]=3+n−12[2n+4]=3+(n−1)(n+2)=3+n2−n+2n−2=n2+n+1
Thus, the kth term of series:
Tk=k2++k+1
Now taking sum, we get
Sn=n∑k=1Tk=n∑k=1(k2++k+1)=n∑k=1k2+n∑k=1k+n∑k=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).
Question 2
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+…+Tn−1+Tn.
Subtracting second expression from the first expression, we have
Sn−Sn=1+4+10+22+46+…+Tn−(1+4+10+22+…+Tn−1+Tn)
⟹0=1+(4−1)+(10−4)+(22−10)+(46−33)+…+(Tn−Tn−1)−Tn.⟹0=1+(3+6+12+24+… up to (n−1) terms )−Tn
Then
Tn=1+(3+6+12+24+… up to (n−1) terms )=1+3(2n−1−1)2−1(∵Sn=a(rn−1)r−1)=1+3(2n−1−1)=1+3⋅2n−1−3=3⋅2n−1−2.
Thus, the kth term of series:
Tk=3⋅2k−1−2
Now taking sum, we get
Sn=n∑k=1Tk=n∑k=1(3⋅2k−1−2)=3n∑k=12k−1−2n∑k=11=3(1+2+22+23+... up to n terms)−2n=3(1)(2n−1)2−1−2n=3(2n−1)−2n
Hence the sum of given series is 3(2n−1)−2n.
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