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Sequences and Series question

2007 · Shift 1 · Q37
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Sequences and Series question

2007 · Shift 1 · Q37

JEE AdvancedMathematicsSequences and SeriesMCQ+3 / −1
Let V r_rr​ denote the sum of the first r terms of an arithmetic progression (A.P.) whose first term is r and the common difference is (2r−12r-12r−1). Let Tr=Vr+1−Vr−2{T_r} = {V_{r + 1}} - {V_r} - 2Tr​=Vr+1​−Vr​−2 and Qr=Tr+1−Tr{Q_r} = {T_{r + 1}} - {T_r}Qr​=Tr+1​−Tr​ for r = 1, 2, ...T r_rr​ is always
  1. A
    an odd number
  2. B
    an even number
  3. C
    a prime number
  4. D
    a composite number
View written solutionFree

Correct answer: A

  1. Write the sum VrV_rVr​ explicitly

For each fixed rrr, we are given an A.P. with:

  • first term =r= r=r
  • common difference =2r−1= 2r-1=2r−1

The sum of the first rrr terms is

Vr=r2[2r+(r−1)(2r−1)].V_r = \frac{r}{2}\left[2r+(r-1)(2r-1)\right].Vr​=2r​[2r+(r−1)(2r−1)].

Now simplify:

(r−1)(2r−1)=2r2−3r+1.(r-1)(2r-1)=2r^2-3r+1.(r−1)(2r−1)=2r2−3r+1.

So,

2r+(r−1)(2r−1)=2r+2r2−3r+1=2r2−r+1.2r+(r-1)(2r-1)=2r+2r^2-3r+1=2r^2-r+1.2r+(r−1)(2r−1)=2r+2r2−3r+1=2r2−r+1.

Hence,

Vr=r2(2r2−r+1).V_r=\frac{r}{2}(2r^2-r+1).Vr​=2r​(2r2−r+1).
  1. Find Tr=Vr+1−Vr−2T_r = V_{r+1}-V_r-2Tr​=Vr+1​−Vr​−2

First,

Vr+1=r+12(2(r+1)2−(r+1)+1).V_{r+1}=\frac{r+1}{2}\left(2(r+1)^2-(r+1)+1\right).Vr+1​=2r+1​(2(r+1)2−(r+1)+1).

Simplify inside:

2(r+1)2−(r+1)+1=2(r2+2r+1)−r−1+1=2r2+3r+2.2(r+1)^2-(r+1)+1=2(r^2+2r+1)-r-1+1=2r^2+3r+2.2(r+1)2−(r+1)+1=2(r2+2r+1)−r−1+1=2r2+3r+2.

Thus,

Vr+1=r+12(2r2+3r+2).V_{r+1}=\frac{r+1}{2}(2r^2+3r+2).Vr+1​=2r+1​(2r2+3r+2).

Now,

Tr=Vr+1−Vr−2=12[(r+1)(2r2+3r+2)−r(2r2−r+1)]−2.T_r=V_{r+1}-V_r-2 =\frac{1}{2}\left[(r+1)(2r^2+3r+2)-r(2r^2-r+1)\right]-2.Tr​=Vr+1​−Vr​−2=21​[(r+1)(2r2+3r+2)−r(2r2−r+1)]−2.

Expand:

(r+1)(2r2+3r+2)=2r3+5r2+5r+2,(r+1)(2r^2+3r+2)=2r^3+5r^2+5r+2,(r+1)(2r2+3r+2)=2r3+5r2+5r+2, r(2r2−r+1)=2r3−r2+r.r(2r^2-r+1)=2r^3-r^2+r.r(2r2−r+1)=2r3−r2+r.

Subtract:

(2r3+5r2+5r+2)−(2r3−r2+r)=6r2+4r+2.(2r^3+5r^2+5r+2)-(2r^3-r^2+r)=6r^2+4r+2.(2r3+5r2+5r+2)−(2r3−r2+r)=6r2+4r+2.

Therefore,

Tr=12(6r2+4r+2)−2=3r2+2r+1−2=3r2+2r−1.T_r=\frac{1}{2}(6r^2+4r+2)-2=3r^2+2r+1-2=3r^2+2r-1.Tr​=21​(6r2+4r+2)−2=3r2+2r+1−2=3r2+2r−1.
  1. Find Qr=Tr+1−TrQ_r = T_{r+1}-T_rQr​=Tr+1​−Tr​

Compute Tr+1T_{r+1}Tr+1​:

Tr+1=3(r+1)2+2(r+1)−1.T_{r+1}=3(r+1)^2+2(r+1)-1.Tr+1​=3(r+1)2+2(r+1)−1.

Expand:

Tr+1=3(r2+2r+1)+2r+2−1=3r2+8r+4.T_{r+1}=3(r^2+2r+1)+2r+2-1=3r^2+8r+4.Tr+1​=3(r2+2r+1)+2r+2−1=3r2+8r+4.

Now,

Qr=Tr+1−Tr=(3r2+8r+4)−(3r2+2r−1)=6r+5.Q_r=T_{r+1}-T_r=(3r^2+8r+4)-(3r^2+2r-1)=6r+5.Qr​=Tr+1​−Tr​=(3r2+8r+4)−(3r2+2r−1)=6r+5.
  1. Determine the nature of QrQ_rQr​

We have

Qr=6r+5.Q_r=6r+5.Qr​=6r+5.

Since 6r6r6r is even, 6r+56r+56r+5 is odd for every positive integer rrr.

So QrQ_rQr​ is always an odd number.

It is not always prime: for example, when r=4r=4r=4,

Q4=6⋅4+5=29Q_4=6\cdot 4+5=29Q4​=6⋅4+5=29

(prime), but when r=2r=2r=2,

Q2=17Q_2=17Q2​=17

(prime), and when r=1r=1r=1,

Q1=11.Q_1=11.Q1​=11.

However, for r=6r=6r=6,

Q6=41,Q_6=41,Q6​=41,

prime again, while for r=8r=8r=8,

Q8=53,Q_8=53,Q8​=53,

and for r=9r=9r=9,

Q9=59.Q_9=59.Q9​=59.

To check composite possibility, take r=5r=5r=5:

Q5=35,Q_5=35,Q5​=35,

which is composite. Thus it is not always prime or always composite.

Hence the only always true statement is that QrQ_rQr​ is odd.

  1. Compare with stored answer

Derived answer: A

Stored correct answer: D

These do not match. The stored answer appears incorrect because

Qr=6r+5,Q_r=6r+5,Qr​=6r+5,

which is always odd, but not always composite.

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