If x_n = frac{2n^2 + n + 1}{2n^2 - 3n + 2},th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $x_n = \frac{2n^2 + n + 1}{2n^2 - 3n + 2}$.Let $f(n) = 2n^2 - 3n + 2$. Then $f(n-1) = 2(n-1)^2 - 3(n-1) + 2 = 2(n^2 - 2n + 1) - 3n + 3 + 2 =

Vedclass · Accepted Answer

$\frac{n(n+3)}{2}$

Vedclass · Answer

(B) Given $x_n = \frac{2n^2 + n + 1}{2n^2 - 3n + 2}$.Let $f(n) = 2n^2 - 3n + 2$. Then $f(n-1) = 2(n-1)^2 - 3(n-1) + 2 = 2(n^2 - 2n + 1) - 3n + 3 + 2 = 2n^2 - 4n + 2 - 3n + 5 = 2n^2 - 7n + 7$. This does not match the denominator directly.Let's re-evaluate: $x_n = \frac{2n^2 + n + 1}{2(n-1)^2 + (n-1) + 1} = \frac{f(n)}{f(n-1)}$ where $f(n) = 2n^2 + n + 1$.Then $\prod_{i=1}^r x_i = \frac{f(1)}{f(0)} \cdot \frac{f(2)}{f(1)} \cdots \frac{f(r)}{f(r-1)} = \frac{f(r)}{f(0)}$.$f(0) = 2(0)^2 + 0 + 1 = 1$. So $\prod_{i=1}^r x_i = 2r^2 + r + 1$.Also,$\sum_{i=1}^r (2i - 1) = r^2$.Thus,the expression becomes $\sum_{r=1}^n [(2r^2 + r + 1) - 2(r^2)] = \sum_{r=1}^n (r + 1)$.$= \sum_{r=1}^n r + \sum_{r=1}^n 1 = \frac{n(n+1)}{2} + n = \frac{n^2 + n + 2n}{2} = \frac{n(n+3)}{2}$.

If $x_n = \frac{2n^2 + n + 1}{2n^2 - 3n + 2}$,then $\sum_{r=1}^n \left[ \left( \prod_{i=1}^r x_i \right) - 2\sum_{i=1}^r (2i - 1) \right]$ is equal to:

Similar Questions

The sum of the first $n$ terms of the series $1^2 + 2.2^2 + 3^2 + 2.4^2 + 5^2 + 2.6^2 + \dots$ is $\frac{n(n + 1)^2}{2}$ when $n$ is even. When $n$ is odd,the sum will be:

If $a, b, c$ are in $G.P.$,then

In a $G.P.$,the first term is $7$,the last term is $448$,and the sum is $889$. Find the common ratio.

Let the sum of the first three terms of an $A.P.$ be $39$ and the sum of its last four terms be $178.$ If the first term of this $A.P.$ is $10,$ then the median of the $A.P.$ is

If the $4^{th}, 7^{th}$ and $10^{th}$ terms of a $G.P.$ are $a, b, c$ respectively,then the relation between $a, b, c$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module