Question

$\rm \: If \: x_n = \dfrac{2 {n}^{2} + n + 1}{2 {n}^{2} - 3n + 2 } \: \: then \: \displaystyle\[ \sum_{ \rm r=1}^{ \rm n} \] \bigg[\bigg( \prod_{i=1}^{r}  \rm x_i \bigg) - 2\sum_{i=1}^{r}(2i - 1) \bigg] = ?$
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Answer 1

Topic :-

Sequence and Series

Given :-

$\sf {x_n=\dfrac{2n^2+n+1}{2n^2-3n+2}}$

To Find :-

$\displaystyle \sf{\sum_{r=1}^{n}\left[ \left( \prod_{i=1}^{r}x_i\right)-2\sum_{i=1}^{r}(2i-1)\right]}$

Solution :-

$\sf {x_i=\dfrac{2i^2+i+1}{2i^2-3i+2}}$

$\displaystyle \sf{\prod_{i=1}^{r}\dfrac{2i^2+i+1}{2i^2-3i+2}}$

We will put values of 'i' one by one,

$\sf {\left(\dfrac{2(1)^2+1+1}{2(1)^2-3(1)+2}\right)\cdot\left(\dfrac{2(2)^2+2+1}{2(2)^2-3(2)+2}\right)\cdots\cdots\left(\dfrac{2r^2+r+1}{2r^2-3r+2}\right)}$

$\sf {\left(\dfrac{\cancel{4}}{1}\right)\cdot\left(\dfrac{\cancel{11}}{\cancel{4}}\right)\cdots\cdots\left(\dfrac{2r^2+r+1}{\cancel{2r^2-3r+2}}\right)}$

$\sf{\dfrac{2r^2+r+1}{1}}$

$\boxed{\displaystyle \sf{\prod_{i=1}^{r}\dfrac{2i^2+i+1}{2i^2-3i+2}=2r^2+r+1}}$

$\displaystyle \sf {\sum_{i=1}^{r}(2i-1)=Sum\:of\:first\:'r'\:odd\:natural\:numbers}$

$\boxed{\displaystyle \sf {\sum_{i=1}^{r}(2i-1)=r^2}}$

$\sf {\left( \because Sum\:of\:first\:'n'\:odd\:natural\:numbers=n^2\right)}$

$\displaystyle \sf{\sum_{r=1}^{n}\left[ \left( \prod_{i=1}^{r}x_i\right)-2\sum_{i=1}^{r}(2i-1)\right]}$

$\displaystyle \sf{\sum_{r=1}^{n}\left[ \left( 2r^2+r+1\right)-2(r^2)\right]}$

$\displaystyle \sf{\sum_{r=1}^{n}\left[2r^2+r+1-2r^2\right]}$

$\displaystyle \sf{\sum_{r=1}^{n}\:[r+1]}$

$\displaystyle \sf{\sum_{r=1}^{n}\:r+\sum_{r=1}^{n}\:1}$

$\displaystyle \sf{\dfrac{n(n+1)}{2}+\sum_{r=1}^{n}\:1}$

$\displaystyle \sf{\left(\because\sum_{r=1}^{n}\:r=Sum\:of\:first\:'n'\:natural\:numbers=\dfrac{n(n+1)}{2}\right)}$

$\displaystyle \sf{\dfrac{n(n+1)}{2}+n}$

$\displaystyle \sf{\left(\because\sum_{r=1}^{n}\:1=Sum\:of\:'1'\:n-times=n\right)}$

$\displaystyle \sf{\dfrac{n^2+n+2n}{2}}$

$\displaystyle \sf{\dfrac{n^2+3n}{2}}$

Answer :-

$\underline{\boxed{\displaystyle \sf{\sum_{r=1}^{n}\left[ \left( \prod_{i=1}^{r}x_i\right)-2\sum_{i=1}^{r}(2i-1)\right]=\dfrac{n^2+3n}{2}}}}$