Sequences And Series Ques 41
Passage Based Problems
Read the following passage and answer the questions.
Passage
Let $V _r$ denotes the sum of the first $r$ terms of an arithmetic progression (AP) whose first term is $r$ and the common difference is $(2 r-1)$. Let $T _r=V _{r+1}-V _r$ and $Q _r=T _{r+1}-T _r$ for $r=1,2, \ldots$
(2007, 8M)
The sum $V _1+V _2+\ldots+V _n$ is
(a) $\frac{1}{12} n(n+1)\left(3 n^{2}-n+1\right)$
(b) $\frac{1}{12} n(n+1)\left(3 n^{2}+n+2\right)$
(c) $\frac{1}{2} n\left(2 n^{2}-n+1\right)$
(d) $\frac{1}{3}\left(2 n^{3}-2 n+3\right)$
Show Answer
Answer:
Correct Answer: 41.(b)
Solution:
Formula:
An Arithmetic Progression (A.P.):
- Here, $V _r=\frac{r}{2}[2 r+(r-1)(2 r-1)]=\frac{1}{2}\left(2 r^{3}-r^{2}+r\right)$
$\therefore \Sigma V _r=\frac{1}{2}\left[2 \Sigma r^{3}-\Sigma r^{2}+\Sigma r\right]$
$=\frac{1}{2} [2 \{\frac{n(n+1)}{2}^{2}\}-\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}]$
$\Rightarrow \frac{n(n+1)}{12}[3 n(n+1)-(2 n+1)+3]$
$=\frac{1}{12} n(n+1)\left(3 n^{2}+n+2\right)$
BETA
