IIT-JEE 2007 · Paper 1 · Mathematics Q58

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MathematicsSingle correct+4 / -1Question 58 of 66

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Paragraph for Question Nos. 58 to 60

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
$T_r=V_{r+1}-V_r-2\quad\text{and}\quad Q_r=T_{r+1}-T_r\quad\text{for }r=1,2,\ldots$

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

\frac1{12}n(n+1)(3n^2-n+1)

\frac1{12}n(n+1)(3n^2+n+2)

\frac12n(2n^2-n+1)

\frac13(2n^3-2n+3)

Answer and worked solution

Official answer: B

The A.P. sum.
$V_r=\frac r2\left[2r+(r-1)(2r-1)\right]=\frac r2\left(2r^2-r+1\right)=\frac{2r^3-r^2+r}2.$

Summing with $\sum r^3=\frac{n^2(n+1)^2}4$ , $\sum r^2=\frac{n(n+1)(2n+1)}6$ , $\sum r=\frac{n(n+1)}2$ :
$\sum_{r=1}^nV_r=\frac{n(n+1)}{12}\left[3n(n+1)-(2n+1)+3\right]=\frac1{12}n(n+1)\left(3n^2+n+2\right).$

(B)

Topics:Special series sums Arithmetic progression

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IIT-JEE 2007 · Paper 1, Mathematics question 58