#1**+8 **

This problem is an altered form of an easier problem, which I will explain first before I'll move on to your problem.

Suppose we want to add the numbers 1 to 10.

Then we could go about this by simply calculating

$$\begin{array}{lcl}

1+1 = 2\\

2+3 = 5\\

5+4 = 9\\

9+5 = 14\\

\mbox{etc.}

\end{array}$$

Since this takes quite a long time, we can also try to find a more logical way to solve this.

Let's try to find pairs which make it easier for us to solve this summation.

$$\begin{array}{lcl}

0+10 = 10\\

1+9 = 10\\

2+8 = 10\\

3+7 = 10\\

4+6 = 10\\

\ \ \ \ \ \ 5 = 5\\

\_\_\_\_\_\_\_\_\_\_\_\_\_ \\

1+...+10 = 55\\

\mbox{Hence the sum of all the numbers between 1 and 10 is 55}\\

\end{array}$$

$$\begin{array}{lcl}

\mbox{Note that we could also write this as } 10\times \frac{10}{2}+\frac{10}{2}\\

\mbox{This again can be rewritten to} \frac{10 \times 10 + 10}{2}\\

\mbox{and } \frac{10 \times 10 + 10}{2} = \frac{10 \times (10+1)}{2}

\end{array}$$

Try to see for yourself that this idea doesn't just hold for the sum of the numbers 1 to 10, but also for numbers 1 to n. where n is any number you want to choose. Therefore

$$1+2+...+n = \frac{n(n+1)}{2}$$

Now find the sum of 1 to 201, the sum of 1 to 100 and then substract the second from the first :D

Then you can find the sum of 1 to 201.

Good luck!

reinout-g Nov 11, 2014

#1**+8 **

Best Answer

This problem is an altered form of an easier problem, which I will explain first before I'll move on to your problem.

Suppose we want to add the numbers 1 to 10.

Then we could go about this by simply calculating

$$\begin{array}{lcl}

1+1 = 2\\

2+3 = 5\\

5+4 = 9\\

9+5 = 14\\

\mbox{etc.}

\end{array}$$

Since this takes quite a long time, we can also try to find a more logical way to solve this.

Let's try to find pairs which make it easier for us to solve this summation.

$$\begin{array}{lcl}

0+10 = 10\\

1+9 = 10\\

2+8 = 10\\

3+7 = 10\\

4+6 = 10\\

\ \ \ \ \ \ 5 = 5\\

\_\_\_\_\_\_\_\_\_\_\_\_\_ \\

1+...+10 = 55\\

\mbox{Hence the sum of all the numbers between 1 and 10 is 55}\\

\end{array}$$

$$\begin{array}{lcl}

\mbox{Note that we could also write this as } 10\times \frac{10}{2}+\frac{10}{2}\\

\mbox{This again can be rewritten to} \frac{10 \times 10 + 10}{2}\\

\mbox{and } \frac{10 \times 10 + 10}{2} = \frac{10 \times (10+1)}{2}

\end{array}$$

Try to see for yourself that this idea doesn't just hold for the sum of the numbers 1 to 10, but also for numbers 1 to n. where n is any number you want to choose. Therefore

$$1+2+...+n = \frac{n(n+1)}{2}$$

Now find the sum of 1 to 201, the sum of 1 to 100 and then substract the second from the first :D

Then you can find the sum of 1 to 201.

Good luck!

reinout-g Nov 11, 2014