Q. 35 - Calculus

Question

Find a formula for each of the sums in Exercises 35, and then use these formulas to calculate each sum for $n=100,n=500$ and $n=1000$.

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

Step-by-Step Solution

Answer

The sum is $\begin{array}{rcl}& & \frac{5n-n^2}{2}\end{array}$.

The sum when $n=100$ is $-4750$.

The sum when $n=500$ is $-123750$.

The sum when $n=1000$ is $-497500$.

Step 1: Given information

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

Step 2: Determine the formula for the given summation.

The sum can be written as:

$\begin{array}{rcl}\sum _{k=1}^n(3-k)& =& 3\sum _{k=1}^{n}1-\sum _{k=1}^{n}k\\ & =& 3n-\frac{n\left(n+1\right)}{2} [\sum _{k=1}^{n}k=\frac{n\left(n+1\right)}{2} \mathrm{and} \sum _{k=1}^{n}1=n ]\\ & =& \frac{6n-n^2-n}{2}\\ & =& \frac{5n-n^2}{2}\end{array}$

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

Substitute $100$ for $n$ in $\begin{array}{rcl}& & \frac{5n-n^2}{2}\end{array}$.

$\begin{array}{rcl}\frac{5\left(100\right)-{\left(100\right)}^2}{2}& =& \frac{500-10000}{2}\\ & =& \frac{-9500}{2}\\ & =& -4750\end{array}$

Substitute $500$ for $n$ in $\begin{array}{rcl}& & \frac{5n-n^2}{2}\end{array}$.

$\begin{array}{rcl}\frac{5\left(500\right)-{\left(500\right)}^2}{2}& =& \frac{2500-250000}{2}\\ & =& -123750\end{array}$

Substitute $1000$ for $n$ in $\begin{array}{rcl}& & \frac{5n-n^2}{2}\end{array}$.

$\begin{array}{rcl}\frac{5\left(1000\right)-{\left(1000\right)}^2}{2}& =& \frac{5000-1000000}{2}\\ & =& -497500\end{array}$

Step 4: Write the conclusion

The formula is $\begin{array}{rcl}& & \frac{5n-n^2}{2}\end{array}$.

The sum when $n=100,500$ and $1000$ is $-4750,-123750$ and $-497500$.