Q. 38 - Calculus

Question

Find a formula for each of the sums in Exercises 38, and then use these formulas to calculate each sum for $n=100,500$ and $1000$

Step-by-Step Solution

Answer

The formula of given summation is $\frac{-4n+n^4+2n^3+n^2}{16}$.

The sum when $n=100$ is $6375600$.

The sum when $n=500$ is $3921890500$.

The sum when $n=1000$ is $62625062250$.

Step 1: Given information

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

Step 2: Determine the formula for the given summation.

The sum can be written as: 

$$\begin{gathered} \sum _{k=1}^n\frac{k^3-1}{4}=\frac{1}{4}\sum _{k=1}^n\left(k^3-1\right) \\ =\frac{1}{4}\left(\sum _{k=1}^n{k}^3-\sum _{k=1}^{n}1\right) \\ =\frac{1}{4}\left(\frac{n^2(n+1{)}^2}{4}-n\right) \\ =\frac{-4n+n^4+2n^3+n^2}{16} \end{gathered}$$

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

Substitute $100$ for $n$ in $\frac{-4n+n^4+2n^3+n^2}{16}$.

$\frac{-4\left(100\right)+{\left(100\right)}^4+2{\left(100\right)}^3+{\left(100\right)}^2}{16}=6375600$

Substitute $500$ for $n$ in $\frac{-4n+n^4+2n^3+n^2}{16}$.

$\frac{-4\left(500\right)+{\left(500\right)}^4+2{\left(500\right)}^3+{\left(500\right)}^2}{16}=3921890500$

Substitute $1000$ for $n$ in $\frac{-4n+n^4+2n^3+n^2}{16}$.

$\frac{-4\left(1000\right)+{\left(1000\right)}^4+2{\left(1000\right)}^3+{\left(1000\right)}^2}{16}=62625062250$

Step 4: Write the conclusion

The formula is $\frac{-4n+n^4+2n^3+n^2}{16}$.

The sum when $n=100,500,$ and $1000$ are $6375600,3921890500,$and $62625062250$.