Question

The value of $cot\left\{{\sum }_{n=1}^{23}co{t}^{-1}(1+{\sum }_{k=1}^{n}2k)\right\}$ is

• A. $\frac{23}{25}$
• B. $\frac{25}{23}$
• C. $\frac{23}{24}$
• D. $\frac{24}{23}$

Answer 1

The correct option is B $\frac{25}{23}$

We have, $cot\left[{\sum }_{n=1}^{23}co{t}^{-1}(1+{\sum }_{k=1}^{n}2k)\right]\Rightarrow cot\left[{\sum }_{n=1}^{23}co{t}^{-1}\right(1+2+4+6+8\dots +2n\left)\right]\Rightarrow cot\left[{\sum }_{n=1}^{23}co{t}^{-1}\{1+n(n+1\left)\right\}\right]\Rightarrow cot\left[{\sum }_{n=1}^{23}ta{n}^{-1}\frac{1}{1+n(n+1)}\right]\Rightarrow cot\left[{\sum }_{n=1}^{23}ta{n}^{-1}\left\{\frac{(n+1)-n}{1+n(n+1)}\right\}\right]\Rightarrow cot\left[{\sum }_{n=1}^{23}\right(ta{n}^{-1}(n+1)-ta{n}^{-1}lnn\left)\right]$
$\Rightarrow cot\left[\right(ta{n}^{-1}2-ta{n}^{-1}1)+(ta{n}^{-1}3-ta{n}^{-1}2)+(ta{n}^{-1}4-ta{n}^{-1}3)+\dots +(ta{n}^{-1}24-ta{n}^{-1}2\left)\right]\Rightarrow cot(ta{n}^{-1}24-ta{n}^{-1}1)\Rightarrow cot\left(ta{n}^{-1}\frac{24-1}{1+24.\left(1\right)}\right)=cot\left(ta{n}^{-1}\frac{23}{25}\right)=cot\left(co{t}^{-1}\frac{25}{23}\right)=\frac{25}{23}$