101+102+...201=?

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!

$$Sn = \frac{(a1+an)*n}{2}

S = the sum of the numbers

a1 = first number

an = last number

n = quantity of numbers

an = a1 + d(n-1) [d = 1 = the common difference of successive members)

201 = 101 - 1(n-1)

n = 101

S101 = [(101+201)*101]/2

S101 = 15251$$

.