Question

$\text{Let}{S}_n={\sum }_{l=1}^n\left(\frac{l^4+l^{3}n+l^2{n}^2+2n^4}{n^5}\right)\text{and}{T}_n={\sum }_{l=0}^{n-1}\left(\frac{l^4+l^{3}n+l^2{n}^2+2n^4}{n^5}\right),(n=1,2,3,...)\text{then}$

• A. $T_n>\frac{167}{60}$
• B. $T_n<\frac{167}{60}$
• C. $S_n>\frac{167}{60}$
• D. $S_n<\frac{167}{60}$

Answer 1

Answer: (B), (C)

$S_n>\frac{167}{60}$

Replace $\frac{l}{n}\to x\text{and}\frac{1}{n}\to dx$, The functoin we get is $f\left(x\right)=x^4+x^3+x^2+2$
$f\left(x\right)$ is increasing in (0,1)
$S_n={\sum }_{l=1}^n\frac{l^4}{n^4}.\frac{1}{n}+\frac{l^3}{n^3}.\frac{1}{n}+\frac{l^2}{n^2}.\frac{1}{n}+\frac{2}{n}$

$(S_n\text{is the sum of all the rectangles}$
$>{\int }_0^1(x^4+x^3+x^2+2\left)dx\right(\because S_n>\text{area under curve})=\frac{167}{60}$
$\Rightarrow S_n>\frac{167}{60}$


$T_n={\sum }_{l=0}^{n-1}\left(\frac{l^4+l^{3}n+l^2{n}^2+2n^4}{n^5}\right)(\because T_n<\text{area under curve})$
$\Rightarrow T_n<{\int }_0^{1}f\left(x\right)dx$
$\Rightarrow T_n<\frac{167}{60}$