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

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(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 =

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

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(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:

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