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 $f(n-1) = 2(n-1)^2 - 3(n-1) + 2 = 2(n^2 - 2n + 1) - 3n + 3 + 2 = 2n^2 - 7n + 7$. This does not match the numerator directly. Let us re-evaluate $x_n = \frac{2n^2 + n + 1}{2(n-1)^2 - 3(n-1) + 2} = \frac{2n^2 + n + 1}{2n^2 - 7n + 7}$.Actually,observing the product $\prod_{i=1}^r x_i$: for $r=1$,$x_1 = \frac{2+1+1}{2-3+2} = \frac{4}{1} = 4$. For $r=2$,$x_1 x_2 = 4 	imes \frac{8+2+1}{8-6+2} = 4 	imes \frac{11}{4} = 11$. For $r=3$,$x_1 x_2 x_3 = 11 	imes \frac{18+3+1}{18-9+2} = 11 	imes \frac{22}{11} = 22$. The product is $2r^2 + r + 1$.The second term is $2 \sum_{i=1}^r (2i - 1) = 2(r^2) = 2r^2$.Thus,the expression inside the summation is $(2r^2 + r + 1) - 2r^2 = r + 1$.Finally,$\sum_{r=1}^n (r + 1) = \sum_{r=1}^n r + \sum_{r=1}^n 1 = \frac{n(n+1)}{2} + n = \frac{n^2 + n + 2n}{2} = \frac{n(n+3)}{2}$.

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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