Question

Assertion :$\frac{1^2}{\left(1\right)\left(3\right)}+\frac{2^2}{\left(3\right)\left(5\right)}+.....+\frac{n^2}{(2n-1)(2n+1)}=\frac{n(n+1)}{2(2n+1)}$ because Reason: $\frac{1}{\left(1\right)\left(3\right)}+\frac{1}{\left(3\right)\left(5\right)}+.....+\frac{1}{(2n-1)(2n+1)}=\frac{1}{2n+1}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

Let $t_r=\frac{r^2}{(2r-1)(2r+1)}$

$\Rightarrow 4t_r=\frac{4r^2-1+1}{(2r-1)(2r+1)}=1+\frac{1}{(2r-1)(2r+1)}=1+\frac{1}{2}(\frac{1}{2r-1}-\frac{1}{2r+1})$

$4{\sum }_{r=1}^n{t}_r=n+\frac{1}{2}{\sum }_{r=1}^n(\frac{1}{2r-1}-\frac{1}{2r+1})=n+\frac{1}{2}(1-\frac{1}{2n+1})=n+\frac{n}{2n+1}=\frac{2n(n+1)}{2n+1}$

$\Rightarrow {\sum }_{r=1}^n{t}_r=\frac{n(n+1)}{2(2n+1)}$

$\therefore$ Statement -1 is true

But statement-2 is false since

$\Rightarrow {\sum }_{r=1}^n\frac{1}{(2r-1)(2r+1)}=\frac{1}{2}{\sum }_{r=1}^n(\frac{1}{2r-1}-\frac{1}{2r+1})=\frac{1}{2}(1-\frac{1}{2n+1})=\frac{n}{2n+1}$