Step-by-Step Solution

1. Express V$_r$ in terms of r

We are given that V$_r$ is the sum of the first r terms of an Arithmetic Progression (A.P.). For this A.P.:

• First term, $a = r$
• Common difference, $d = 2r - 1$
• Number of terms, $n = r$

The formula for the sum of the first n terms of an A.P. is $S_n = {n \over 2}[2a + (n - 1)d]$. Substituting the given values to find V$_r$:

$V_r = {r \over 2}[2(r) + (r - 1)(2r - 1)]$

Now, let's simplify the expression inside the brackets:

$V_r = {r \over 2}[2r + (2r^2 - r - 2r + 1)]$ $V_r = {r \over 2}[2r + 2r^2 - 3r + 1]$ $V_r = {r \over 2}[2r^2 - r + 1]$ $V_r = r^3 - {1 \over 2}r^2 + {1 \over 2}r$

Note: The definitions of T$_r$ and Q$_r$ are not needed to find the sum of V$_r$, so they can be ignored for this specific question.

2. Calculate the sum $\sum_{r=1}^{n} V_r$

The question asks for the sum V$_1$ + V$_2$ + ... + V$_n$, which can be written as a summation:

$S_n = \sum_{r=1}^{n} V_r = \sum_{r=1}^{n} \left(r^3 - {1 \over 2}r^2 + {1 \over 2}r\right)$

We can separate the summation term by term:

$S_n = \sum_{r=1}^{n} r^3 - {1 \over 2}\sum_{r=1}^{n} r^2 + {1 \over 2}\sum_{r=1}^{n} r$

3. Apply standard summation formulas

We use the following standard formulas for the sum of powers of the first n natural numbers:

• $\sum_{r=1}^{n} r = {n(n + 1) \over 2}$
• $\sum_{r=1}^{n} r^2 = {n(n + 1)(2n + 1) \over 6}$
• $\sum_{r=1}^{n} r^3 = \left[{n(n + 1) \over 2}\right]^2 = {n^2(n + 1)^2 \over 4}$

Substitute these formulas into the expression for S$_n$:

$S_n = {n^2(n + 1)^2 \over 4} - {1 \over 2} \left({n(n + 1)(2n + 1) \over 6}\right) + {1 \over 2} \left({n(n + 1) \over 2}\right)$ $S_n = {n^2(n + 1)^2 \over 4} - {n(n + 1)(2n + 1) \over 12} + {n(n + 1) \over 4}$

4. Simplify the expression for S$_n$

To simplify, we find a common denominator, which is 12, and factor out the common term $n(n+1)$:

$S_n = {n(n + 1) \over 12} \left[ 3n(n + 1) - (2n + 1) + 3 \right]$

Now, simplify the expression inside the square brackets:

$S_n = {n(n + 1) \over 12} [3n^2 + 3n - 2n - 1 + 3]$ $S_n = {n(n + 1) \over 12} [3n^2 + n + 2]$

5. Compare with the given options

The final expression is $S_n = {1 \over {12}}n(n + 1)(3{n^2} + n + 2)$. Let's check the options:

A: ${1 \over {12}}n(n + 1)(3{n^2} - n + 1)$ B: ${1 \over {12}}n(n + 1)(3{n^2} + n + 2)$ C: ${1 \over 2}n(2{n^2} - n + 1)$ D: ${1 \over 3}(2{n^3} - 2n + 3)$

Our result matches option B.

Verification (Optional)

Let's check for n=1:

• $V_1 = 1^3 - {1 \over 2}(1)^2 + {1 \over 2}(1) = 1 - {1 \over 2} + {1 \over 2} = 1$.
• The sum is $S_1 = V_1 = 1$.
• Using the formula from option B: $S_1 = {1 \over {12}}(1)(1 + 1)(3(1)^2 + 1 + 2) = {1 \over {12}}(1)(2)(6) = {12 \over 12} = 1$. This matches.

Let's check for n=2:

• $V_2 = 2^3 - {1 \over 2}(2)^2 + {1 \over 2}(2) = 8 - {4 \over 2} + 1 = 8 - 2 + 1 = 7$.
• The sum is $S_2 = V_1 + V_2 = 1 + 7 = 8$.
• Using the formula from option B: $S_2 = {1 \over {12}}(2)(2 + 1)(3(2)^2 + 2 + 2) = {1 \over {12}}(2)(3)(12 + 2 + 2) = {6 \over 12}(16) = {1 \over 2}(16) = 8$. This also matches.