Sequence and Series: The sum V_1+V_2+cdots+V_n is | JEE Advanced 2007 Paper 1

Comprehension · Questions 1–1

Read the passage given below and answer the questions that follow.

Let $V_r$ denote the sum of the first $r$ terms of an arithmetic progression (A.P.) whose first term is $r$ and the common difference is $(2r-1)$. Let $T_r = V_{r+1}-V_r-2$ and $Q_r = T_{r+1}-T_r$ for $r=1,2,\ldots$

JEE Advanced 2007 Paper 1Question 1

The sum $V_1+V_2+\cdots+V_n$ is

Sequence and Series · Standard Sums (Σ formulas) · Arithmetic Progression

$\frac{1}{12}n(n+1)(3n^2-n+1)$$\frac{1}{12}n(n+1)(3n^2+n+2)$✓ Correct answer$\frac{1}{2}n(2n^2-n+1)$$\frac{1}{3}(2n^3-2n+3)$

Solution

Formulas used. (1) From the passage, $V_r$ is the sum of the first $r$ terms of an A.P. with first term $r$ and common difference $2r-1$: $V_r=\frac r2\left[2r+(r-1)(2r-1)\right]$(2) $\sum r^3=\frac{n^2(n+1)^2}{4}$, $\ \sum r^2=\frac{n(n+1)(2n+1)}{6}$, $\ \sum r=\frac{n(n+1)}{2}$Step 1 - Simplify $V_r$. $$V_r=\frac r2\left(2r^2-r+1\right)=r^3-\frac{r^2}{2}+\frac r2 .$$ Step 2 - Sum. $$\sum_{r=1}^{n}V_r=\frac{n^2(n+1)^2}{4}-\frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}{4}.$$ Step 3 - Take $\frac{n(n+1)}{12}$ common. $$\frac{n(n+1)}{12}\left[3n(n+1)-(2n+1)+3\right]=\frac{n(n+1)}{12}\left(3n^2+n+2\right).$$ Answer: (B)

Sequence and Series - JEE Mathematics question (JEE Advanced 2007 Paper 1)

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