Question

Let Vr denote the sum of the first r terms of an arithmetic progression, whose first term is r and the common difference is given by \[\left( {2r - {\text{ }}1} \right)\]. Let \[T_r=V_{r+1}-V_r-2\] and \[Q_r=T_{r+1}-T_r\] for \[r{\text{ }} = {\text{ }}1,2 \ldots .\] The sum \[V_1+V_2+......+V_n\] is
(a) \[\left( {1/12} \right)n\left( {n + 1} \right)\left( {3{n^2} + n + 2} \right)\]
(b) \[\left( {1/2n} \right)\left( {2{n^2} - n + 1} \right)\]
(c) \[\left( {1/3} \right)\left( {2{n^3} - 2n + 3} \right)\]
(d) None of these

Answer 1

Hint:The first step is to substitute the given values in the formula \[{S_n} = \dfrac{n}{2}[2a + (n - 1)d]\] and simplify it. To proceed further, use the basic formulas for the sum of n numbers, the sum of squares of n numbers, and the sum of cubes of n numbers. Simplify the obtained equations to get the value of \[V_1+V_2+......+V_n\].

Formula Used:
  \[{S_n} = \dfrac{n}{2}[2a + (n - 1)d]\]
\[\sum {n^3} = {\left[ {\dfrac{{n(n + 1)}}{2}} \right]^2}\]
\[\sum {n^2} = \left[ {\dfrac{{n(n + 1)(2n + 1)}}{2}} \right]\]
 \[\sum n = \left[ {\dfrac{{n(n + 1)}}{2}} \right]\]

Complete step by step Solution:
To begin with, let’s recall the general formula to find the sum on n terms of arithmetic equation i.e.
 \[{S_n} = \dfrac{n}{2}[2a + (n - 1)d]\]--- eq (1)
Given in the question Vr represents the sum of the first r terms of an arithmetic progression (AP), where first term (a) is denoted by r and the common difference (d) is represented by \[\left( {2r - {\text{ }}1} \right)\]. Substitute these values in eq (1)
\[{V_r} = \dfrac{r}{2}[2r + (r - 1)(2r - 1)]\]
Simplify the above equation and opening the brackets.
\[{V_r} = \dfrac{r}{2}(2r + 2{r^2} - r - 2r + 1)\]
Cancel out the like terms and multiply the remaining terms.
\[{V_r} = \dfrac{1}{2}(2{r^3} - {r^2} + r)\]
Now, find the summation of Vr i.e. \[\sum {V_r}\]
..-- eq (2)
We know the general formulas for the sum of n numbers, sum of square of n numbers, and sum of cube of n numbers which are given by
\[\sum {n^3} = {\left[ {\dfrac{{n(n + 1)}}{2}} \right]^2}\], \[\sum {n^2} = \left[ {\dfrac{{n(n + 1)(2n + 1)}}{2}} \right]\], \[\sum n = \left[ {\dfrac{{n(n + 1)}}{2}} \right]\]
Use the above formulas in eq (2)
\[\sum {V_r} = \dfrac{1}{2}\left( {2 \times \dfrac{{{r^2}{{(r + 1)}^2}}}{4} - \dfrac{{r(r + 1)(2r + 1)}}{6} + \dfrac{{r(r + 1)}}{2}} \right)\]
Take out \[(r + 1)\]as common term and simplify the remaining terms.
\[\sum {V_r} = \dfrac{{r(r + 1)}}{4}\left[ {\dfrac{{3{r^2} + 3r - 2r - 1 + 3}}{3}} \right]\]
Simplify it further to get,
\[\sum {V_r} = \dfrac{{r(r + 1)}}{{12}}\left[ {\dfrac{{3{r^2} + r + 2}}{3}} \right]\]
Replacing r by n in the above equation, we get
\[\Sigma {V_n} = \dfrac{1}{{12}}n(n + 1)(3{n^2} + n + 2)\]
Therefore, the summation of \[V_1+V_2+......+V_n\] is \[\dfrac{1}{{12}}n(n + 1)(3{n^2} + n + 2)\]

Hence, the correct option is 1.

Note: The sum of n terms of an arithmetic progression (AP) is given by: \[{S_n} = \dfrac{n}{2}[2a + (n - 1)d]\]. The general formulas should be kept in mind while solving such problems. Care must be taken while simplifying or opening the brackets to avoid mistakes.