JEE Challenger
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MathMedAdv 2007 • Paper 1

Sum of Terms of Sequence Formed by Arithmetic Progressions

Comprehension Passage

Let VrV_r denote the sum of the first rr terms of an arithmetic progression (A.P.) whose first term is rr and the common difference is (2r−1)(2r - 1). Let
Tr=Vr+1−Vr−2 and Qr=Tr+1−Tr for r=1,2,…T_r = V_{r+1} - V_r - 2 \text{ and } Q_r = T_{r+1} - T_r \text{ for } r = 1, 2, \dots

The sum V1+V2+⋯+VnV_1 + V_2 + \dots + V_n is

Options

A

112n(n+1)(3n2−n+1)\frac{1}{12}n(n + 1)(3n^2 - n + 1)

B

112n(n+1)(3n2+n+2)\frac{1}{12}n(n + 1)(3n^2 + n + 2)

Correct
C

12n(2n2−n+1)\frac{1}{2}n(2n^2 - n + 1)

D

13(2n3−2n+3)\frac{1}{3}(2n^3 - 2n + 3)

Topics & Concepts

Sequences and SeriesArithmetic ProgressionSum of Series

Step-by-Step Solution

To find the sum V1+V2+⋯+VnV_1 + V_2 + \dots + V_n, we first determine the expression for VrV_r, which denotes the sum of the first rr terms of an arithmetic progression (A.P.).

For a given rr:

  • First term of the A.P., a=ra = r
  • Common difference, d=2r−1d = 2r - 1
  • Number of terms = rr

The formula for the sum of rr terms of an A.P. is:
Vr=r2[2a+(r−1)d]V_r = \frac{r}{2} \left[ 2a + (r - 1)d \right]

Substituting a=ra = r and d=2r−1d = 2r - 1:
Vr=r2[2r+(r−1)(2r−1)]V_r = \frac{r}{2} \left[ 2r + (r - 1)(2r - 1) \right]
Vr=r2[2r+2r2−3r+1]V_r = \frac{r}{2} \left[ 2r + 2r^2 - 3r + 1 \right]
Vr=r2(2r2−r+1)V_r = \frac{r}{2} \left( 2r^2 - r + 1 \right)
Vr=r3−12r2+12rV_r = r^3 - \frac{1}{2}r^2 + \frac{1}{2}r

Now, we calculate the sum ∑r=1nVr\sum_{r=1}^n V_r:
∑r=1nVr=∑r=1nr3−12∑r=1nr2+12∑r=1nr\sum_{r=1}^n V_r = \sum_{r=1}^n r^3 - \frac{1}{2}\sum_{r=1}^n r^2 + \frac{1}{2}\sum_{r=1}^n r

Using the standard summation formulas:

  • ∑r=1nr=n(n+1)2\sum_{r=1}^n r = \frac{n(n + 1)}{2}
  • ∑r=1nr2=n(n+1)(2n+1)6\sum_{r=1}^n r^2 = \frac{n(n + 1)(2n + 1)}{6}
  • ∑r=1nr3=n2(n+1)24\sum_{r=1}^n r^3 = \frac{n^2(n + 1)^2}{4}

Substituting these formulas into the sum:
∑r=1nVr=n2(n+1)24−12⋅n(n+1)(2n+1)6+12⋅n(n+1)2\sum_{r=1}^n V_r = \frac{n^2(n + 1)^2}{4} - \frac{1}{2} \cdot \frac{n(n + 1)(2n + 1)}{6} + \frac{1}{2} \cdot \frac{n(n + 1)}{2}

Factoring out n(n+1)12\frac{n(n + 1)}{12}:
∑r=1nVr=n(n+1)12[3n(n+1)−(2n+1)+3]\sum_{r=1}^n V_r = \frac{n(n + 1)}{12} \left[ 3n(n + 1) - (2n + 1) + 3 \right]

Simplifying the term inside the bracket:
3n(n+1)−(2n+1)+3=3n2+3n−2n−1+3=3n2+n+23n(n + 1) - (2n + 1) + 3 = 3n^2 + 3n - 2n - 1 + 3 = 3n^2 + n + 2

Therefore, the sum is:
V1+V2+⋯+Vn=112n(n+1)(3n2+n+2)V_1 + V_2 + \dots + V_n = \frac{1}{12}n(n + 1)(3n^2 + n + 2)

This corresponds to Option (B).

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