Q. 39 - Calculus

Question

Find a formula for each of the sums in Exercises 39, and then use these formulas to calculate each sum for $n=100,500$ and $1000$

$\sum _{k=1}^n\frac{k^3-1}{n^4}$

Step-by-Step Solution

Answer

The formula of the given summation is $\frac{\left(n-1\right)\left(n^2+3n+4\right)}{4n^3}$.

The sum when $n=100$ is $0.255024$.$0.255024$

The sum when $n=500$ is $\hspace{0.17em}0.251000992$.

The sum when $n=1000$ is $0.250500249$.$0.250500249$.

Step 1: Given information

The given summation is $\sum _{k=1}^n\frac{k^3-1}{n^4}$.

Step 2: Determine the formula for the given summation.

The sum can be written as:

$$\begin{gathered} \sum _{k=1}^n\frac{k^3-1}{n^4}=\frac{1}{n^4}\sum _{k=1}^n\left(k^3-1\right) \\ =\frac{1}{n^4}\left(\sum _{k=1}^n{k}^3-\sum _{k=1}^{n}1\right) \\ =\frac{1}{n^4}\left(\frac{n^2(n+1{)}^2}{4}-n\right) \\ =\frac{\left(n-1\right)\left(n^2+3n+4\right)}{4n^3} \end{gathered}$$

Step 3: Evaluate the sum for n = 100 , 500 and 1000

Substitute $100$ for $n$ in $\frac{\left(n-1\right)\left(n^2+3n+4\right)}{4n^3}$.

$\frac{\left(100-1\right)\left({\left(100\right)}^2+3\left(100\right)+4\right)}{4{\left(100\right)}^3}=0.255024$

Substitute $500$ for $n$ in $\frac{\left(n-1\right)\left(n^2+3n+4\right)}{4n^3}$.

$\frac{\left(500-1\right)\left({\left(500\right)}^2+3\left(500\right)+4\right)}{4{\left(500\right)}^3}=\hspace{0.17em}0.251000992$

Substitute $1000$ for $n$ in $\frac{\left(n-1\right)\left(n^2+3n+4\right)}{4n^3}$.

$\frac{\left(1000-1\right)\left({\left(1000\right)}^2+3\left(1000\right)+4\right)}{4{\left(1000\right)}^3}=0.250500249$

Step 4: Write the conclusion

The formula is $\frac{\left(n-1\right)\left(n^2+3n+4\right)}{4n^3}$.

The sum when $n=100,500$ and $1000$ are $0.255024,0.251000992$ and $0.250500249$.