Question

If $I_1=\underset{n\to \infty }{lim}\sum _{r=1}^{2n}\frac{n}{r^2+3rn+2n^2}$

and $I_2=\underset{n\to \infty }{lim}\sum _{r=1}^{2n}\frac{n}{2r^2+3rn+n^2}$

then the value of $2I_1-I_2$ is equal to?

• A. $\ln \frac{4}{3}$
• B. $\ln \frac{3}{4}$
• C. $\ln \frac{3}{2}$
• D. $0$

Answer 1

Answer: (C)

$\ln \frac{3}{2}$

$I_1={\int }_0^2\frac{dx}{x^2+3x+2}$
$I_2={\int }_0^2\frac{dx}{2x^2+3x+1}$

Substitute $x=\frac{t}{2}$
$dx=\frac{dt}{2}$

$I_2={\int }_0^2\frac{dt}{2(2\frac{t^2}{4}+\frac{3t}{2}+1)}={\int }_0^2\frac{dt}{t^2+3t+2}=I_1$

$I_1={\int }_0^2\frac{dx}{{(x+\frac{3}{2})}^2-{\left(\frac{1}{2}\right)}^2}={\ln \left(\frac{x+1}{x+2}\right)|}_0^2$

$=\ln \frac{3}{4}-\ln \frac{1}{2}=\ln \frac{3}{2}$
Ans: C