Q. 36 - Calculus

Question

Find a formula for each of the sums in Exercises 36, and then use these formulas to calculate each sum for $n=100,500$ and $n=1000$

$\sum _{k=1}^n\left(k^3-10k^2+2\right)$

Step-by-Step Solution

Answer

The formula of the given summation is $\frac{3n^4-34n^3-57n^2+4n}{12}$.

The sum when $n=100$ is $24716191700$.

The sum when $n=500$ is $15269646000$.

The sum when $n=1000$ is $247161917000$.

Step 1: Given information

The given summation is $\sum _{k=1}^n\left(k^3-10k^2+2\right)$.

Step 2: Determine the formula for the given summation

The sum can be written as:

$$\begin{gathered} \sum _{k=1}^n\left(k^3-10k^2+2\right)=\sum _{k=1}^n{k}^3-10\sum _{k=1}^n{k}^2+2\sum _{k=1}^{n}1 \\ =\frac{n^2(n+1{)}^2}{4}-10\left(\frac{n(n+1)(2n+1)}{6}\right)+2n \left[\sum _{k=1}^n{k}^3=\frac{n^2(n+1{)}^2}{4}, \sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}, \sum _{k=1}^{n}1=n\right] \\ =\frac{3n^4-34n^3-57n^2+4n}{12} \end{gathered}$$

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

Substitute $100$ for $n$ in $\frac{3n^4-34n^3-57n^2+4n}{12}$.

$\frac{3{\left(100\right)}^4-34{\left(100\right)}^3-57{\left(100\right)}^2+4\left(100\right)}{12}=24716191700$

Substitute $500$ for $n$ in $\frac{3n^4-34n^3-57n^2+4n}{12}$.

$\frac{3{\left(500\right)}^4-34{\left(500\right)}^3-57{\left(500\right)}^2+4\left(500\right)}{12}=15269646000$

Substitute $1000$ for $n$ in $\frac{3n^4-34n^3-57n^2+4n}{12}$.

$\frac{3{\left(1000\right)}^4-34{\left(1000\right)}^3-57{\left(1000\right)}^2+4\left(1000\right)}{12}=247161917000$

Step 4: Write the conclusion

The formula is $\frac{3n^4-34n^3-57n^2+4n}{12}$.

The sum when $n=100,500$ and $1000$ is $24716191700, 15269646000, \mathrm{and} 247161917000$