Q. 37 - Calculus

Question

Find a formula for each of the sums in Exercises 37, and then use these formulas to calculate each sum for $n=100,500$ and $1000$.

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

Step-by-Step Solution

Answer

The formula is $\begin{array}{rcl}& & \frac{2n^3+9n^2+13n}{6}\end{array}$.

The sum when $n=100$ is .$348550$

The sum when $n=500$ is $\begin{array}{rcl}& & 42042750\end{array}$.

The sum when $n=1000$ is $\begin{array}{rcl}& & 334835500\end{array}$.

Step 1: Given information

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

Step 2: Determine the formula of the given summation

The sum can be written as:

$\begin{array}{rcl}\sum _{k=3}^n(k+1{)}^2& =& \sum _{k=3}^n(k^2+2k+1)\\ & =& \sum _{k=3}^n{k}^2+2\sum _{k=3}^{n}k+\sum _{k=3}^{n}1\\ & =& \left(\frac{n(n+1)(2n+1)}{6}\right)+2\frac{n(n+1)}{2}+n \left[\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}, \sum _{k=1}^{n}k=\frac{n(n+1)}{2}, \sum _{k=1}^{n}1=n\right]\\ & =& \frac{2n^3+9n^2+13n}{6}\end{array}$

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

Substitute $100$ for $n$ in $\begin{array}{rcl}& & \frac{2n^3+9n^2+13n}{6}\end{array}$.

$\begin{array}{rcl}& & \frac{2{\left(100\right)}^3+9{\left(100\right)}^2+13\left(100\right)}{6}\end{array}=348550$

Substitute $500$ for $n$ in $\begin{array}{rcl}& & \frac{2n^3+9n^2+13n}{6}\end{array}$.

$\begin{array}{rcl}\frac{2{\left(500\right)}^3+9{\left(500\right)}^2+13\left(500\right)}{6}& =& 42042750\end{array}$

Substitute $1000$ for $n$ in $\begin{array}{rcl}& & \frac{2n^3+9n^2+13n}{6}\end{array}$.

$\begin{array}{rcl}\frac{2{\left(1000\right)}^3+9{\left(1000\right)}^2+13\left(1000\right)}{6}& =& 334835500\end{array}$

Step 4: Write the conclusion

The formula is $\begin{array}{rcl}& & \frac{2n^3+9n^2+13n}{6}\end{array}$.

The sum when  $n=100,500$ and $1000$ is $348550, 42042750$ and $334835500$.