Question

Assertion :If $t_r=\frac{1^2+2^2+3^2+...+r^2}{1^3+2^3+...+r^3}$ and $S_n=\sum _{r=1}^n{(-1)}^r.t_r,$ then $\underset{n\to \infty }{lim}{S}_n=\frac{2}{3}$ Reason: $1^2+2^2+3^2+...+r^2=\frac{r(r+1)(2r+1)}{6}$ and $1^3+2^3+3^3+...+r^3={\left(\frac{r(r+1)}{2}\right)}^2$

• 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 D Assertion is incorrect but Reason is correct

$t_r=\frac{1^2+2^2+3^2+...+r^2}{1^3+2^3+3^3+...+r^3}=\frac{r(r+1)(2r+1)}{6}.{\left(\frac{2}{r(r+1)}\right)}^2=\frac{2}{3}(\frac{1}{r}+\frac{1}{r+1})$

$\therefore S_n=\frac{2}{3}[-(1+\frac{1}{2})+(\frac{1}{2}+\frac{1}{3})-(\frac{1}{3}+\frac{1}{4})+...\pm (\frac{1}{n}+\frac{1}{n+1})]$

$=\frac{2}{3}(-1\pm \frac{1}{n+1})$

$\therefore \underset{n\to \infty }{lim}{S}_n=-\frac{2}{3}$