The sum V₁+V₂ + .. + Vₙ is [JEE Adv 2007 P1]

Correct Answer is : 112n(n+1)(3n2+n+2)1/12 n(n+1)(3n²+n+2)121n(n+1)(3n2+n+2)

The sum V₁+V₂ + .. + Vₙ is [JEE Adv 2007 P1]

Correct Answer is : 112n(n+1)(3n2+n+2)1/12 n(n+1)(3n²+n+2)121n(n+1)(3n2+n+2)

Test Index

IIT JEE Advanced 2007 Paper 1

Paragraph Type

This section contains 2 paragraphs. Based on each paragraph, there are 3 questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Question 58–60:

Let

V_r

done 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 , ...

Section: Mathematics

Question : 58 of 66

Marks: +1, -0

The sum

V_1+V_2

+ .. +

V_n

[JEE Adv 2007 P1]

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

Validate

Solution:

We have

v_r

\frac{r}{2} [2r+(r-1)(2r-1)]

\frac{1}{2}(2r^3-r^2+r)

Therefore,

\sum v_r

\sum \frac{1}{2} (2r^3-r^2+r)

\sum r^3 - \frac{1}{2} \sum r^2 + \frac{1}{2} \sum r

\frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{12} + \frac{n(n+1)}{4}

\frac{n(n+1)}{12}

[3n (n + 1) - (2n + 1) + 3]

\frac{1}{12}

n (n + 1) (

3n^2

+ n + 2)

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